Question 5.10: Acceleration of Two Objects Connected by a Cord A ball of ma...

Conceptualize Imagine the objects in Figure 5.16 in motion. If m_2 moves down the incline, then m_1 moves upward. Because the objects are connected by a cord (which we assume does not stretch), their accelerations have the same magnitude. Notice the normal coordinate axes in Figure 5.16b for the ball and the “tilted” axes for the block in Figure 5.16c. Just as we chose the positive direction to be different for each of the objects in Example 5.9, we are free to choose entirely different coordinate axes for the two objects here.

Categorize We can identify forces on each of the two objects and we are looking for an acceleration, so we categorize the objects as particles under a net force. For the block, this model is only valid for the x′ direction. In the y′ direction, we apply the particle in equilibrium model because the block does not accelerate in that direction.

Analyze Consider the free-body diagrams shown in Figures 5.16b and 5.16c.

Apply Newton’s second law in the y direction to the ball, choosing the upward direction as positive:

(1)   \Sigma F_y=T-m_1 g=m_1 a_y=m_1 a

For the ball to accelerate upward, it is necessary that T \gt m_1g. In Equation (1), we replaced a_y with a because the acceleration has only a y component.

For the block, we have chosen the x′ axis along the incline as in Figure 5.15c. For consistency with our choice for the ball, we choose the positive x′ direction to be down the incline.

Apply the particle under a net force model to the block in the x′ direction and the particle in equilibrium model in the y′ direction:

(2)    \sum F_{x^{\prime}}=m_2 g \sin \theta-T=m_2 a_{x^{\prime}}=m_2 a

(3)   \sum F_{y^{\prime}}=n-m_2 g \cos \theta=0

In Equation (2), we replaced a_{x^{\prime}} with a because the two objects have accelerations of equal magnitude a.

Solve Equation (1) for T:

(4)   T=m_1(g+a)

Substitute this expression for T into Equation (2):

Solve for a:

(5)   a=\left(\frac{m_2 \sin \theta-m_1}{m_1+m_2}\right) g

Substitute this expression for a into Equation (4) to find T:

(6)   T=\left[\frac{m_1 m_2(\sin \theta+1)}{m_1+m_2}\right] g

Finalize The block accelerates down the incline only if m_2 \sin \theta>m_1. If m_1>m_2 \sin \theta, the acceleration is up the incline for the block and downward for the ball. Also notice that the result for the acceleration, Equation (5), can be interpreted as the magnitude of the net external force acting on the ball-block system divided by the total mass of the system; this result is consistent with Newton’s second law.