Dynamics solutions set 1

From the handout:

1. acceleration of block = .824 m/s2.
2. Tension in cable = 21000 N, lift force = 72500 N
3. Skydiver acceleration = 5.95 m/s^2 directed downwards
4. Slowing car: acceleration = 1.2 m/s^2, braking force = 1200 N
5. Woman must accelerate DOWN at 2.727 m/s^2
6. a) 980 N, b) 1010 N c) 965 N d) 980 N e) 0 N
7. a) 539 N b) 539 N c) 717 N d) 361 N e) 0 N


Book problems:
Question 11. Although the 2 kg rock has twice the gravity force, it also has twice the mass. By Newton's 2nd law, the acceleration must therefore be the same.

Problems:
12. T = 12960 N

13. accelerates DOWN at 3.5 m/s^2

15. If the thief is able to accelerate down the rope, the tension required to prevent him from falling will be less than his weight, and the "rope" will stay together.

Using the same logic as problem 5 from the handout, the acceleration must be downwards at 2.221 m/s^2 minimum for the rope not to break.

16. accelerates DOWN at .25 g, or at 2.45 m/s^2.

17. Using g = 9.8 m/s^2, the maximum upwards acceleration is .557 m/s^2. Using g = 10 m/s^2, the maximum acceleration is .357 m/s^2.

30.
a) top cord tension = 58.8 N, bottom cord tension is 29.4 N
b) top cord tension = 68.4 N, bottom cord is 34.2 N

39.
a) Since the 40 N force is the minimum force required to get the box moving, the static friction force F_s = F_s,max = mu*F_n. Using a FBD and observing the box is in equilibrium until it starts to move, mu = F/mg = 0.8.

b) F - mu*mg = ma, so mu = (F - ma)/mg = 0.73

44. (After drawing a FBD of course!...) From Newton's 2nd, a = mu*g, and delta_x = -v_o^2/(2*mu*g) = 4.082 meters.

Statics with Friction

From handout:

Problem 1: Fn = 400 N, Fk = 173.2 N, mu = .433
Problem 2: Fn = 250 N, Fk = 173.2 N, mu = .693

Systems Problem (We will talk about this first thing tomorrow. There are some subtle points to discuss on how to do these types of problems.

Mass M = mu*M = 2 kg

Book Problems:
Q22. Since friction force is proportional to the normal force, putting a force against the box and perpendicular to the wall increases the normal force, which then increases the friction force.

Problems:
38. Friction force = 102.9 N. If mu = 0, there wouldn't be any friction force, and the only magnitude of F that could lead to constant velocity across the floor would be zero!

43. 6.667 kg

59. You should obtain that tan(phi) = mu. Thus phi = arc_tan(0.6) = 31.0 degrees.

63.
T = mg = 26.46 N,
Fk = mg - mg*sin(theta) = mu * Fn
Fn = mg cos(theta)

mu = 0.637

Problem 54. The clown must pull with a force of 308.3 N.

Static Equilibrium problems

Some solutions!

Problems:
6. (I will use g = 10m/s^2 to make things easier to calculate.)
a)weight = 200 N, normal force 200 N
b) Table exerts 300 N, bottom box exerts a normal force ON the top box of 100 N.

28.
a)FBD should include F at 45 degrees below the horizontal to0 the right, a friction force directed to the left, and the weight (mg) directed downwards.
b) Friction force F = 88 N cos(45) = 62.2 N
c) Fn = 88 N sin 45 + (14.5 kg)(9.8 m/s^2) = 204.3 N

78.


page 267:
12. T1 (diagonal cord) = 3920 N, T2 = 3395 N
13. right cord = 176.9 N, left cord = 234.8 N
16. tension = 48.2 N. The tension is so great because of the small angle. 2T sin(theta) = mg. Since theta is small, T is large compared with the weight mg.

Review Solutions

Solutions to the MC questions

Pages 1 - 3:
1 E
2 C
3 E
4 C
5 A
6 D
7 B
8 D
9 B
10 D

Page 4:

Page 5 - 6
1. B
2. E

4. E

6. D
7. D
8. C

More Projectile motion!

The written solutions to the handout problems are located at this link.

Robot problem:
Acceleration = -0.5 m/s^2, displacement = 27 meters9.0 meters.

We can look at the other solutions to the handout problems tomorrow in class. I will post others when I can.

Projectile Motion

Question 11. Since the vertical component is zero at the top of the parabolic path, and the horizontal component is the same throughout, the minimum speed occurs at the point of maximum height.

Problems:

20. Cliff is 44.1 meters tall, lands 4.8 meters from the base of the cliff.

22.
The pebbles have a constant horizontal velocity, and zero vertical velocity when they hit the window.

Since the vertical displacement is 8.0 meters, the initial vertical speed is 12.52 m/s. This means (from v = v0 + a*delta_t) that the time to hit the window is 1.277 seconds.

Since delta_x = 9.0 meters, the horizontal component of velocity = 9.0 m / 1.277 seconds = 7.05 m/s. This is the speed of the pebbles hitting the window.

24. The drop takes 3.38 seconds using just the vertical direction. The horizontal speed (which stays constant) is 45 m / 3.38 seconds = 13.31 m/s.

26. It takes 1.228 seconds for the ball to reach its maximum, so it takes double this, or 2.456 seconds to fall back down to the ground.

27. The drop takes 1.621 seconds (since delta_x and vx are given.) Voy = 0 since it is thrown horizontally. The height delta_y of the building is then 12.88 meters.

28. You can set up an equation using the equation for delta_y and time.

-2.2 m = (14 m/s)sin(40)*t - .5*(9.8 m/s^2)*t^2

There are two solutions (times) when this is true, but only the positive root matters: this is at 2.055 seconds. The horizontal displacement is given by (14 m/s)*cos(40)*2.055 seconds = 22.04 meters.

Problem 5 from the handout:
vo = D/(2H/g)^1/2

WP3. Both bullets hit the ground at the same time.

Vector Kinematics

Two important things to keep in mind - keep components separate from each other in a kinematics problem. Second, don't subsitute values until you are ready to get an answer. Keeping things algebraic makes things much cleaner.

Handout Solutions:
5.
a) delta_t = v₀ sin(theta) / g
b) delta_y = (v₀ sin(theta))^2/2g

6. displacement is 65.9 km at [E 15.6 degrees N][E 17.0 degrees N]

7. x-component: 27.19 m/s, y-component: 12.67 m/s

8. 43.01 m/s at 54.46 degrees above the x-axis

9.

10. 60 mi/h 34.28 mi/h

11.
a) 1.72 seconds
b) 14.58 meters
c) double the answer in (a) of 3.44 seconds
d) 124.7 meters

12. Check your notes for these definitions.

Book problems:

Question 10. No - the magnitudes may be the same, but the directions are different. This means the velocities are not the same.

9.
a) West component: 785 km/h* sin(38.5 degrees) = 488.7 km/h (Note that this would be negative according to our usual coordinate system of positive to the right.)
North component: 785 km/h* cos(38.5 degrees) = 614.3 km/h
b)
West: 1466.1 km
North: 1842.9 km

16.
a) Vertical component of acceleration = 3.8 m/s^2 * sin(30) = 1.9 m/s^2
b) delta_y = v0y*delta_t + 1/2*a_y*delta_t^2, but since she starts from rest, v0_y = 0. The delta_y is the elevation change of 335 meters.

Thus delta_t = (2*delta_y/a_y)^1/2 = 18.77 seconds