1. A load of bricks with mass m 1 = 15.2 k g hangs from one end of a rope that p

Question

1. A load of bricks with mass m1 = 15.2kg hangs from one end of a rope that passes over a small, frictionless pulley. A counterweight of mass m2 = 28.8kg is suspended from the other end of the rope, as shown in the figure. The system is released from rest. Use g = 9.80m/s2 for the magnitude of the acceleration

(a) What is the magnitude of the upward acceleration of the load of bricks?

(b) What is the tension in the rope while the load is moving?

2. You are lowering two boxes, one on top of the other, down the ramp shown in the figure by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 12.0

cm/s . The coefficient of kinetic friction between the ramp and the lower box is 0.403, and the coefficient of static friction

(a) What force do you need to exert to accomplish this?

(b) What is the magnitude of the friction force on the upper box?

3. An airplane flies in a loop (a circular path in a vertical plane) of radius 140

m . The pilot's head always points toward the center of the loop. The speed of the airplane is not constant; the airplane goes slowest at the top of the loop and fastest at the bottom.

(a) At the top of the loop, the pilot feels weightless. What is the speed of the airplane at this point?

(b) At the bottom of the loop, the speed of the airplane is 210

km/h . What is the apparent weight of the pilot at this point? His true weight is 800N .

Explanation / Answer

1a)

For the bricks...

T-mg = ma

T = (15.2)(a) + (15.2)(9.8)

T = 15.2a + 148.96

For the counterweight

mg - T = ma

(28.8)(9.8) - 15.2a - 148.96 = (28.8)a

133.28 = 44a

a = 3.03 m/s^2

1b)

T = (15.2)(3.03) + (15.2)(9.8)

T = 195 N

2a)

The angle is found from tan(angle) = 4.75/2.5

Angle = 27.8 degrees

Fa + Ff = Fwsin(angle)

Fa + umgcos(angle) = mgsin(angle)

Fa + (.403)(80)(9.8)(cos 27.8) = (80)(9.8)(sin 27.8)

Fa = 86.2 N

2b) Ff = umgcos(angle)

Ff = (.792)(32)(9.8)(cos 27.8)

Ff = 220 N

3a)

At the top, mv^2/r = mg (mass cancels)

v^2 = rg

v^2 = (140)(9.8)

v = 37 m/s

3b)

At the bottom, Apparant weight = mv^2/r + mg

m = 800/9.8 = 81.63 kg

v = 210 km/hr = 58.33 m/s

Apparant weight = (81.63)(58.33)^2/140 + (81.63)(9.8)

Apparant Weight = 2784 N

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