Physics 106 Mechanics Lecture 08 Wenda Cao NJIT

Physics 106: Mechanics Lecture 08 Wenda Cao NJIT Physics Department

Static Equilibrium q Important announcement q Equilibrium and static equilibrium q Static equilibrium conditions n n Net external force must equal zero Net external torque must equal zero q Solving static equilibrium problems March 10, 2011

Static and Dynamic Equilibrium q Equilibrium implies the object is at rest (static) or its center of mass moves with a constant velocity (dynamic) q 106 deals only with the special case in which linear and angular velocities are equal to zero, called “static equilibrium” : v. CM = 0 and w = 0 q Examples n n n Book on table Puck sliding on ice in a constant velocity Ceiling fan – off Ceiling fan – on Ladder leaning against wall (foot in groove) March 10, 2011

Conditions for Equilibrium The first condition of equilibrium is a statement of translational equilibrium q The net external force on the object must equal zero q q It states that the translational acceleration of the object’s center of mass must be zero March 10, 2011

Conditions for Equilibrium The secondition of equilibrium is a statement of rotational equilibrium q The net external torque on the object must equal zero q It states the angular acceleration of the object to be zero q This must be true for any axis of rotation q March 10, 2011

Conditions for Equilibrium q The n If the object is modeled as a particle, then this is the only condition that must be satisfied q The n net force equals zero net torque equals zero This is needed if the object cannot be modeled as a particle q These conditions describe the rigid objects in equilibrium analysis model March 10, 2011

Equilibrium Equations q Equation 1: q Equation 2: q We will restrict the applications to situations in which all the forces lie in the xy plane q There are three resulting equations March 10, 2011

A) Find the magnitude of the upward force n exerted by the support on the board. B) Find where the father should sit to balance the system at rest. March 10, 2011

Axis of Rotation q The net torque is about an axis through any point in the xy plane q Does it matter which axis you choose for calculating torques? q NO. The choice of an axis is arbitrary q If an object is in translational equilibrium and the net torque is zero about one axis, then the net torque must be zero about any other axis q We should be smart to choose a rotation axis to simplify problems March 10, 2011

B) Find where the father should sit to balance the system at rest. Rotation axis O P Rotation axis P O March 10, 2011

Center of Gravity q The torque due to the gravitational force on an object of mass M is the force Mg acting at the center of gravity of the object q If g is uniform over the object, then the center of gravity of the object coincides with its center of mass q If the object is homogeneous and symmetrical, the center of gravity coincides with its geometric center March 10, 2011

Problem-Solving Strategy 1 q q q q Draw sketch, decide what is in or out the system Draw a free body diagram (FBD) Show and label all external forces acting on the object Indicate the locations of all the forces Establish a convenient coordinate system Find the components of the forces along the two axes Apply the first condition for equilibrium Be careful of signs March 10, 2011

Horizontal Beam Example q A uniform horizontal beam with a length of l = 8. 00 m and a weight of Wb = 200 N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of = 53 with the beam. A person of weight Wp = 600 N stands a distance d = 2. 00 m from the wall. Find the tension in the cable as well as the magnitude and direction of the force exerted by the wall on the beam. March 10, 2011

Horizontal Beam Example, 2 q Analyze n n Draw a free body diagram Use the pivot in the problem (at the wall) as the pivot n n This will generally be easiest Note there are three unknowns (T, R, q) March 10, 2011

Horizontal Beam Example, 3 q The forces can be resolved into components in the free body diagram q Apply the two conditions of equilibrium to obtain three equations q Solve for the unknowns March 10, 2011

Problem-Solving Strategy 2 q Choose a convenient axis for calculating the net torque on the object n q Choose an origin that simplifies the calculations as much as possible n q A force that acts along a line passing through the origin produces a zero torque Be careful of sign with respect to rotational axis n n n q Remember the choice of the axis is arbitrary positive if force tends to rotate object in CCW negative if force tends to rotate object in CW zero if force is on the rotational axis Apply the secondition for equilibrium March 10, 2011

Horizontal Beam Example, 3 March 10, 2011

Problem-Solving Strategy 3 q q q The two conditions of equilibrium will give a system of equations Solve the equations simultaneously Make sure your results are consistent with your free body diagram If the solution gives a negative for a force, it is in the opposite direction to what you drew in the free body diagram Check your results to confirm March 10, 2011

Example 3: A uniform beam, of length L and mass m = 1. 8 kg, is at rest with its ends on two scales (see figure). A uniform block, with mass M = 2. 7 kg, is at rest on the beam, with its center a distance L / 4 from the beam's left end. What do the scales read? Solve as equilibrium system March 10, 2011

1: 2: 3: Choose axis at O O From 2: March 10, 2011

Example 4: A safe whose mass is M = 430 kg is hanging by a rope from a boom with dimensions a = 1. 9 m and b = 2. 5 m. The boom consists of a hinged beam and a horizontal cable that connects the beam to a wall. The uniform beam has a mass m of 85 kg; the mass of the cable and rope Fv are negligible. Tc Tr mg Fh (a) What is the tension Tc in the cable; i. e. , what is the magnitude of the force Tc on the beam from the horizontal cable? (b) What is the force at the hinge? March 10, 2011

Choose axis at hinge March 10, 2011

Example 7: Rock Climber In the figure, a rock climber with mass m = 55 kg rests during a “chimney climb, ” pressing only with her shoulders and feet against the walls of a fissure of width w = 1. 0 m. Her center of mass is a horizontal distance d = 0. 20 m from the wall against which her shoulders are pressed. The coefficient of static friction between her shoes and the wall is m 1 = 1. 1, and between her shoulders and the wall it is m 2 = 0. 70. To rest, the climber wants to minimize her horizontal push on the walls. The minimum occurs when her feet and her shoulders are both on the verge of sliding. m 2 m 1 (a) What is the minimum horizontal push on the walls? (b) What should the vertical distance h be between the shoulders and feet, in order for her to remain stationary? March 10, 2011

(a) What is the minimum horizontal push on the walls? (b) What should the vertical distance h be between the shoulders and feet, in order for her to remain stationary? O Choose axis at “O” w = 1. 0 m March 10, 2011