Presentation is loading. Please wait.

Presentation is loading. Please wait.

Homework Solutions for Dynamics. Homework 8 Solutions.

Similar presentations


Presentation on theme: "Homework Solutions for Dynamics. Homework 8 Solutions."— Presentation transcript:

1 Homework Solutions for Dynamics

2 Homework 8 Solutions

3 4. Without friction, the only horizontal force is the tension. We apply Newtons second law to the car: F = ma; F T = (1050 kg)(1.20 m/s 2 ) = 1.26 x 10 3 N.

4

5

6

7 18. With down positive, we write ·F = ma from the force diagram for the skydivers: mg – F R = ma; (a) Before the parachute opens, we have mg –.25mg = ma, which gives a = +g = 7.4 m/s2 (down). (b) Falling at constant speed means the acceleration is zero, so we have mg – F R = ma = 0, which gives F R = mg = (120.0 kg)(9.80 m/s2) = 1176 N.

8 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.

9

10 Homework 9 Solutions

11 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 = v jump 2 + 2(– g)h; 0 = v jump 2 + 2(– 9.80 m/s 2 )(0.80 m), which gives v jump = 3.96 m/s. We can find the acceleration required to achieve this velocity during the crouch from v jump 2 = v 0 2 + 2a(y – y 0 ); (3.96 m/s) 2 = 0 + 2a(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: F N – mg = ma; F N – (66 kg)(9.80 m/s 2 ) = (66 kg)(39.2 m/s 2 ), which gives FN = 3.2 x 10 3 N. From Newtons third law, the person will exert an equal and opposite force on the Ground: 3.2 x 10 3 N downward.

12 21. (a) We find the velocity just before striking the ground from v 1 2 = v 0 2 + 2(– g)h; v 1 2 = 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 1 2 + 2a(y – y 0 ); 0 = (9.4 m/s) 2 + 2a(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 ) – F legs = (45 kg)(– 63 m/s 2 ), which gives F legs = 3.3 x 10 3 N up

13 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.

14

15 Homework 11 Solutions

16

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

18 32. (a) Because the speed is constant, the acceleration is zero. We write ·F = ma from the force diagram: F T + F T – mg = ma = 0, which gives 2F T = mg = 1/2(65 kg)(9.80 m/s 2 ) = 3.2 x 10 2 N. (b) Now we have: F T + F T – mg = ma; 2(1.10)(mg) – mg = ma, which gives a = 0.10g = 0.10(9.80 m/s 2 ) = 0.98 m/s 2

19 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?

20

21 Homework 12 Solutions

22

23

24

25 Do Now: On an icy day,, you worry about parking your car in your driveway, which has an incline of 12 o. Your neighbor Ralphs driveway has an incline of 9.0 o and Bonnies 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?

26

27 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.

28


Download ppt "Homework Solutions for Dynamics. Homework 8 Solutions."

Similar presentations


Ads by Google