Incline and Friction Examples Physics 6 A Prepared

Incline and Friction Examples Physics 6 A Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Friction is a force that opposes the motion of surfaces that are in contact with each other. We will consider 2 types of friction in this class: KINETIC Friction – for surfaces that are in motion (sliding) STATIC Friction – for surfaces at rest Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Friction is a force that opposes the motion of surfaces that are in contact with each other. We will consider 2 types of friction in this class: KINETIC Friction – for surfaces that are in motion (sliding) STATIC Friction – for surfaces at rest The formulas are very similar – each one has a “coefficient of friction” (µ) that determines how much of the Normal force is translated into friction force. Crucial distinction – kinetic friction will be a constant force, while static friction will be just strong enough to keep the surfaces from slipping Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Friction is a force that opposes the motion of surfaces that are in contact with each other. We will consider 2 types of friction in this class: KINETIC Friction – for surfaces that are in motion (sliding) STATIC Friction – for surfaces at rest The formulas are very similar – each one has a “coefficient of friction” (µ) that determines how much of the Normal force is translated into friction force. Crucial distinction – kinetic friction will be a constant force, while static friction will be just strong enough to keep the surfaces from slipping Here are the formulas: See – friction is FUN! Static friction will have a maximum value. If you push any harder the surfaces will slip and you get kinetic friction instead! Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force. a) How hard do you have to push to get the box moving? b) How far will the box travel if you push for 3 seconds? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Normal force b) How far will the box travel if you push for 3 seconds? First we draw a diagram of the forces. friction x Fpush weight Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Normal force b) How far will the box travel if you push for 3 seconds? First we draw a diagram of the forces. friction To get the box moving we have to push hard enough to overcome static friction. So we need to find the maximum force of static friction. x Fpush weight Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Normal force b) How far will the box travel if you push for 3 seconds? First we draw a diagram of the forces. friction To get the box moving we have to push hard enough to overcome static friction. So we need to find the maximum force of static friction. x Fpush weight The coefficient is given, but we need to find the normal force. We can use the y-direction forces to find it. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Normal force b) How far will the box travel if you push for 3 seconds? First we draw a diagram of the forces. friction To get the box moving we have to push hard enough to overcome static friction. So we need to find the maximum force of static friction. x Fpush weight The coefficient is given, but we need to find the normal force. We can use the y-direction forces to find it. Now we can calculate the maximum friction force. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Normal force b) How far will the box travel if you push for 3 seconds? First we draw a diagram of the forces. friction To get the box moving we have to push hard enough to overcome static friction. So we need to find the maximum force of static friction. x Fpush weight The coefficient is given, but we need to find the normal force. We can use the y-direction forces to find it. Now we can calculate the maximum friction force. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Normal force b) How far will the box travel if you push for 3 seconds? First we draw a diagram of the forces. friction To get the box moving we have to push hard enough to overcome static friction. So we need to find the maximum force of static friction. x Fpush weight The coefficient is given, but we need to find the normal force. We can use the y-direction forces to find it. Now we can calculate the maximum friction force. This is how hard we have to push to get the box moving (ok, maybe we push with a force of 588. 0000001 N) Now that we have the answer for part a) how do we do part b)? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Normal force b) How far will the box travel if you push for 3 seconds? friction x Fpush Part b) is really a kinematics problem. To find the answer we need the acceleration of the box. Then we can go back to chapter 2 and use one of our kinematics formulas. We can write down Newton’s 2 nd law for the x-direction: weight What type of friction do we have? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Normal force b) How far will the box travel if you push for 3 seconds? friction x Fpush Part b) is really a kinematics problem. To find the answer we need the acceleration of the box. Then we can go back to chapter 2 and use one of our kinematics formulas. We can write down Newton’s 2 nd law for the x-direction: weight The box is moving, so kinetic friction Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Normal force b) How far will the box travel if you push for 3 seconds? friction x Fpush Part b) is really a kinematics problem. To find the answer we need the acceleration of the box. Then we can go back to chapter 2 and use one of our kinematics formulas. We can write down Newton’s 2 nd law for the x-direction: weight Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Normal force b) How far will the box travel if you push for 3 seconds? friction x Fpush Part b) is really a kinematics problem. To find the answer we need the acceleration of the box. Then we can go back to chapter 2 and use one of our kinematics formulas. We can write down Newton’s 2 nd law for the x-direction: weight Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Normal force b) How far will the box travel if you push for 3 seconds? friction x Fpush Part b) is really a kinematics problem. To find the answer we need the acceleration of the box. Then we can go back to chapter 2 and use one of our kinematics formulas. We can write down Newton’s 2 nd law for the x-direction: weight Now that we have the acceleration we can use our kinematics formula: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Normal force b) How far will the box travel if you push for 3 seconds? friction x Fpush Part b) is really a kinematics problem. To find the answer we need the acceleration of the box. Then we can go back to chapter 2 and use one of our kinematics formulas. We can write down Newton’s 2 nd law for the x-direction: weight Now that we have the acceleration we can use our kinematics formula: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (at an angle of 30° below horizontal) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Let’s do a more difficult version of this problem. This time the push is not horizontal, but down at an angle. Normal force friction x Fpush Think about what will change if the push is not horizontal. weight Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (at an angle of 30° below horizontal) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Let’s do a more difficult version of this problem. This time the push is not horizontal, but down at an angle. Normal force friction x Fpush Think about what will change if the push is not horizontal. Because the push is downward, the Normal force must increase to compensate. This creates more friction than before. weight This might seem like a minor difference, but it makes the algebra much more involved. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (at an angle of 30° below horizontal) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Let’s do a more difficult version of this problem. This time the push is not horizontal, but down at an angle. Normal force friction x Fpush Think about what will change if the push is not horizontal. Because the push is downward, the Normal force must increase to compensate. This creates more friction than before. weight This might seem like a minor difference, but it makes the algebra much more involved. Before we can write the formulas, The pushing force needs to be broken into components. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (at an angle of 30° below horizontal) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Let’s do a more difficult version of this problem. This time the push is not horizontal, but down at an angle. Think about what will change if the push is not horizontal. Because the push is downward, the Normal force must increase to compensate. This creates more friction than before. Normal force x Fpush, x friction Fpush, y weight This might seem like a minor difference, but it makes the algebra much more involved. Before we can write the formulas, The pushing force needs to be broken into components. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (at an angle of 30° below horizontal) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Normal force Now we can write Newton’s 2 nd law for each direction: Fpush, x friction x Fpush, y weight Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (at an angle of 30° below horizontal) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Normal force Now we can write Newton’s 2 nd law for each direction: Fpush, x friction x Fpush, y weight As in the previous problem, we set the acceleration to 0 because the box is not going to move unless we push harder than the maximum static friction. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (at an angle of 30° below horizontal) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Normal force Now we can write Newton’s 2 nd law for each direction: Fpush, x friction x Fpush, y weight At this point we have 2 equations with 2 unknowns (N and Fpush). We can solve for N in the y-equation, then substitute into the x-equation. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (at an angle of 30° below horizontal) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Normal force Now we can write Newton’s 2 nd law for each direction: Fpush, x friction x Fpush, y weight At this point we have 2 equations with 2 unknowns (N and Fpush). We can solve for N in the y-equation, then substitute into the x-equation. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (at an angle of 30° below horizontal) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Normal force Now we can write Newton’s 2 nd law for each direction: Fpush, x friction x Fpush, y weight Now it is basic algebra to solve for Fpush. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example – pushing a box across the floor There is a 100 kg box at rest on a horizontal floor. The coefficients of friction are µk=0. 5 and µs=0. 6. You want to push it across the room, so you push (at an angle of 30° below horizontal) until it starts to slide. Once it starts to move, you keep pushing with constant force. y a) How hard do you have to push to get the box moving? Normal force Now we can write Newton’s 2 nd law for each direction: Fpush, x friction x Fpush, y weight Now it is basic algebra to solve for Fpush. As you can see, the extra downward component of the push makes it much harder to move the box. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500 kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. (ignore friction) θ Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500 kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. (ignore friction) We can draw a diagram of forces for this problem. Sisyphus will push up the incline, but the weight of the boulder will be straight down, so we will split that one into components. θ Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500 kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. (ignore friction) We can draw a diagram of forces for this problem. Sisyphus will push up the incline, but the weight of the boulder will be straight down, so we will split that one into components. Normal FSisyphus θ Wboulder, downhill θ Wboulder Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500 kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. (ignore friction) We can draw a diagram of forces for this problem. Sisyphus will push up the incline, but the weight of the boulder will be straight down, so we will split that one into components. For Sisyphus to be able to push the boulder up the hill, his force must be at least equal to the downhill component of the boulder’s weight. We can write down the formula like this: Normal FSisyphus θ Wboulder, downhill θ Wboulder Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500 kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. (ignore friction) We can draw a diagram of forces for this problem. Sisyphus will push up the incline, but the weight of the boulder will be straight down, so we will split that one into components. For Sisyphus to be able to push the boulder up the hill, his force must be at least equal to the downhill component of the boulder’s weight. We can write down the formula like this: Normal FSisyphus θ Wboulder, downhill θ Wboulder Now all we need is FSisyphus – How do we find that? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500 kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. (ignore friction) We can draw a diagram of forces for this problem. Sisyphus will push up the incline, but the weight of the boulder will be straight down, so we will split that one into components. For Sisyphus to be able to push the boulder up the hill, his force must be at least equal to the downhill component of the boulder’s weight. We can write down the formula like this: Normal FSisyphus θ Wboulder, downhill θ Wboulder Sisyphus’ force can be found from the given information. He can lift 500 kg, so multiplying by g, his force is 4900 N. The angle is given as 20°, so we can plug in to find our answer. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500 kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. (ignore friction) We can draw a diagram of forces for this problem. Sisyphus will push up the incline, but the weight of the boulder will be straight down, so we will split that one into components. For Sisyphus to be able to push the boulder up the hill, his force must be at least equal to the downhill component of the boulder’s weight. We can write down the formula like this: Normal FSisyphus θ Wboulder, downhill θ Wboulder Sisyphus’ force can be found from the given information. He can lift 500 kg, so multiplying by g, his force is 4900 N. The angle is given as 20°, so we can plug in to find our answer. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500 kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0. 3. Same as the last problem, but now with added friction! θ Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500 kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0. 3. This time we will have to include a friction force in our diagram. Which direction should it point? θ Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500 kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0. 3. This time we will have to include a friction force in our diagram. Which direction should it point? DOWNHILL, opposing Sisyphus. W Notice that this time we have labeled an axis system with the xdirection pointing uphill, and the y-direction pointing directly away from the incline. This way we can write down two force formulas. y Fnormal x FSisyphus ƒstatic θ Wboulder, y boulder, x θ Wboulder Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500 kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0. 3. This time we will have to include a friction force in our diagram. Which direction should it point? DOWNHILL, opposing Sisyphus. W Notice that this time we have labeled an axis system with the xdirection pointing uphill, and the y-direction pointing directly away from the incline. This way we can write down two force formulas. y Fnormal x FSisyphus ƒstatic θ Wboulder, y boulder, x θ Wboulder Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500 kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0. 3. This time we will have to include a friction force in our diagram. Which direction should it point? DOWNHILL, opposing Sisyphus. W Notice that this time we have labeled an axis system with the xdirection pointing uphill, and the y-direction pointing directly away from the incline. This way we can write down two force formulas. y Fnormal x FSisyphus ƒstatic θ Wboulder, y boulder, x θ Wboulder In this type of problem we need all the forces to balance out. Even though we want Sisyphus to be able to lift the boulder, we want to be just on the borderline between when the boulder moves and when it doesn’t. Thus we want to be in equilibrium to find the maximum weight. Equilibrium means zero acceleration. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500 kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0. 3. This time we will have to include a friction force in our diagram. Which direction should it point? DOWNHILL, opposing Sisyphus. W Notice that this time we have labeled an axis system with the xdirection pointing uphill, and the y-direction pointing directly away from the incline. This way we can write down two force formulas. y Fnormal x FSisyphus ƒstatic θ Wboulder, y boulder, x θ Wboulder Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500 kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0. 3. This time we will have to include a friction force in our diagram. Which direction should it point? DOWNHILL, opposing Sisyphus. W Notice that this time we have labeled an axis system with the xdirection pointing uphill, and the y-direction pointing directly away from the incline. This way we can write down two force formulas. y Fnormal x FSisyphus ƒstatic θ Wboulder, y boulder, x θ Wboulder Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500 kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0. 3. This time we will have to include a friction force in our diagram. Which direction should it point? DOWNHILL, opposing Sisyphus. W Notice that this time we have labeled an axis system with the xdirection pointing uphill, and the y-direction pointing directly away from the incline. This way we can write down two force formulas. y Fnormal x FSisyphus ƒstatic θ Wboulder, y boulder, x θ Wboulder We are assuming Sisyphus can push with a maximum force of 4900 N. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500 kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0. 3. This time we will have to include a friction force in our diagram. Which direction should it point? DOWNHILL, opposing Sisyphus. W Notice that this time we have labeled an axis system with the xdirection pointing uphill, and the y-direction pointing directly away from the incline. This way we can write down two force formulas. y Fnormal x FSisyphus ƒstatic θ Wboulder, y boulder, x θ Wboulder Here’s where we use our triangles Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500 kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0. 3. This time we will have to include a friction force in our diagram. Which direction should it point? DOWNHILL, opposing Sisyphus. W Notice that this time we have labeled an axis system with the xdirection pointing uphill, and the y-direction pointing directly away from the incline. This way we can write down two force formulas. y Fnormal x FSisyphus ƒstatic θ Wboulder, y boulder, x θ Wboulder Now we can combine our equations by substituting for Fnormal in the x equation. We also know the coefficient of friction. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500 kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0. 3. This time we will have to include a friction force in our diagram. Which direction should it point? DOWNHILL, opposing Sisyphus. W Notice that this time we have labeled an axis system with the xdirection pointing uphill, and the y-direction pointing directly away from the incline. This way we can write down two force formulas. y Fnormal x FSisyphus ƒstatic θ Wboulder, y boulder, x θ Wboulder The final step is just a bit of algebra. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB