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Chapter 5: Problem 29

In Fig. 5-38, object \(B\) weighs \(94.0 \mathrm{lb}\) and object \(A\) weighs \(29.0 \mathrm{lb} .\) Between object \(B\) and the plane the coefficient of static friction is \(0.56\) and the coefficient of kinetic friction is 0.25. (a) Find the acceleration of the system if \(B\) is initially at rest. (b) Find the acceleration if \(B\) is moving up the plane. ( \(c\) ) What is the acceleration if \(B\) is moving down the plane? The plane is inclined by \(42.0^{\circ}\).

Short Answer

Expert verified
The acceleration of the system when \(B\) is at rest, moving up, and moving down the plane can be computed by performing the steps as shown above.

Step by step solution

01

Calculate the weight components

Split the weight of object \(B\) into two components: one parallel to the plane and another perpendicular to the plane. The weight components are given by \(W_{B\parallel} = W_B \sin(\theta)\) and \(W_{B\perp} = W_B \cos(\theta)\) where \(W_B = 94.0\, lb\) is the weight of object \(B\) and \(\theta = 42.0^{\circ}\) is the angle of inclination.
02

Determine the net force

The net force acting on the system when object \(B\) is at rest is equal to the difference between the force due to \(B\) moving down the incline and the static frictional force. The static frictional force can be calculated by multiplying the normal force (which is equal to \(W_{B\perp}\)) by the coefficient of static friction. Thus, \(F_{net} = W_{B\parallel} - \mu_s W_{B\perp}\) where \(\mu_s = 0.56\) is the coefficient of static friction.
03

Compute the acceleration for question (a)

The acceleration of the system can be found using Newton's second law \(F = ma\). Here, \(m\) is the total mass of the system, which is the sum of the weights of objects \(A\) and \(B\) divided by gravity \(g\). But since we are asked in 'lb' units, it is not necessary to divide by gravity. So, \(a = F_{net} / (W_A + W_B)\) where \(W_A = 29.0\, lb\) is the weight of object \(A\).
04

Calculate the acceleration for question (b)

If \(B\) is moving up the plane, the kinetic friction will act in the same direction as \(W_{B\parallel}\). The net force is now \(F_{net} = W_{B\parallel} + \mu_k W_{B\perp} - W_A\) where \(\mu_k = 0.25\) is the coefficient of kinetic friction. The acceleration is found using the same procedure as in step 3.
05

Determine the acceleration for question (c)

If \(B\) is moving down the plane, the kinetic friction acts in the opposite direction to \(W_{B\parallel}\). Now, \(F_{net} = W_{B\parallel} - \mu_k W_{B\perp} - W_A\). Again, calculate the acceleration as in step 3.

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