Homework Solutions for Dynamics Homework 8 Solutions 4

Homework Solutions for Dynamics

Homework 8 Solutions

4. Without friction, the only horizontal force is the tension. We apply Newton’s second law to the car: F = ma; FT = (1050 kg)(1. 20 m/s 2) = 1. 26 x 103 N.

18. With down positive, we write ·F = ma from the force diagram for the skydivers: mg – FR = ma; (a) Before the parachute opens, we have mg –. 25 mg = ma, which gives a = +g = 7. 4 m/s 2 (down). (b) Falling at constant speed means the acceleration is zero, so we have mg – FR = ma = 0, which gives FR = mg = (120. 0 kg)(9. 80 m/s 2) = 1176 N.

Do Now The cable supporting a 2100 kg elevator has a maximum strength of 21, 750 N. What maximum upward acceleration can it give the elevator without breaking? Begin with a free body diagram.

Homework 9 Solutions

20. We find the velocity necessary for the jump from the motion when the person leaves the ground to the highest point, where the velocity is zero: v 2 = vjump 2 + 2(– g)h; 0 = vjump 2 + 2(– 9. 80 m/s 2)(0. 80 m), which gives vjump = 3. 96 m/s. We can find the acceleration required to achieve this velocity during the crouch from vjump 2 = v 02 + 2 a(y – y 0); (3. 96 m/s)2 = 0 + 2 a(0. 20 m – 0), which gives a = 39. 2 m/s 2. Using the force diagram for the person during the crouch, we can write ·F = ma: FN – mg = ma; FN – (66 kg)(9. 80 m/s 2) = (66 kg)(39. 2 m/s 2), which gives FN = 3. 2 x 103 N. From Newton’s third law, the person will exert an equal and opposite force on the Ground: 3. 2 x 103 N downward.

21. (a) We find the velocity just before striking the ground from v 12 = v 02 + 2(– g)h; v 12 = 0 + 2(9. 80 m/s 2)(4. 5 m), which gives v 1 = 9. 4 m/s. (b) We can find the average acceleration required to bring the person to rest from v 2 = v 12 + 2 a(y – y 0); 0 = (9. 4 m/s)2 + 2 a(0. 70 m – 0), which gives a = – 63 m/s 2. Using the force diagram for the person during the crouch, we can write ·F = ma: mg – Flegs = ma; (45 kg)(9. 80 m/s 2) – Flegs = (45 kg)(– 63 m/s 2), which gives Flegs = 3. 3 x 103 N up

Do Now A 75 kg thief wants to escape from a 3 rd story jail window. Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this ‘rope’ to escape? Give quantitative evidence for your answer. Start with a FBD.

Homework 11 Solutions

Note: m 1 g in the lower diagram should read m 2 g

32. (a) Because the speed is constant, the acceleration is zero. We write ·F = ma from the force diagram: FT + FT – mg = ma = 0, which gives 2 FT = mg = 1/2(65 kg)(9. 80 m/s 2) = 3. 2 x 102 N. (b) Now we have: FT’ + FT’ – mg = ma; 2(1. 10)(mg) – mg = ma, which gives a = 0. 10 g = 0. 10(9. 80 m/s 2) = 0. 98 m/s 2

Do Now An elevator in a tall building is allowed to reach a maximum speed of 3. 5 m/s going down. What must the tension be in the cable to stop this elevator over a distance of 3. 0 m if the elevator has a mass of 1300 kg including its occupants?

Homework 12 Solutions

Do Now: On an icy day, , you worry about parking your car in your driveway, which has an incline of 12 o. Your neighbor Ralph’s driveway has an incline of 9. 0 o and Bonnie’s driveway across the street has one of 6. 0 o. The coefficient of static friction between tire rubber and ice is 0. 15. Which driveway(s) will be safe to park in?

Do Now A small block of mass m is given an initial speed v 0 up a ramp inclined at angle θ to the horizontal. It travels a distance d up the ramp and comes to rest. a) Determine a formula for the coefficient of kinetic friction between the block and ramp.