Problem 4 (5 pt.) A 5.0-kg box on a horizontal, frictionless surface is attached

Question

Problem 4 (5 pt.) A 5.0-kg box on a horizontal, frictionless surface is attached to a 2.0-kg weight of by a thin, light wire that passes over a pulley (the wire turns the pulley without slipping or stretching.) The pulley has the shape of a uniform solid disk of mass 3.0 kg radius R about an axis through its center is 1-(1/2) MR?. The box is pulled to the left with a constant force 40 N and moves to the left. and radius 0.10 m. The moment of inertia of the disk of mass M and Draw the free body diagrams for the box, the weight, and the pulley. Write the equations of motion for the box (horizontal), the weight (vertical), and the pulley (rotational). (Three equations) Use these equations to determine the acceleration of the box. a) b) c)

Explanation / Answer

Let's call the horizontal tension Th and the vertical tension Tv.

fbd for the 5 kg mass gives us
Fnet = ma = 5kg * a = Th
Dropping units, we have Th = 5a

fbd for 2 kg mass gives us
Fnet = ma = 2kg * a = mg - Tv = 2kg * 9.8m/s² - Tv
Tv = 19.6 - 2a

fbd for pulley gives us
net torque = (Tv - Th)*r = I = ½mr²(a/r) = ½mra r cancels
Tv - Th = ½ma = ½ * 3kg * a = 1.5kg * a
Plug in for Tv and Th:
19.6 - 2a - 5a = 1.5a
19.6 = 8.5a
a = 2.3 m/s²

Th = 5kg * 2.3m/s² = 11.5 N

Tv = 19.6N - 2kg*2.3m/s² = 15 N

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