A ball of mass m1 and a block of mass m2 are attached by a lightweight cord that passes over a frictionless pulley of negligible mass, as in Figure 5.15a. The block lies on a frictionless incline of angle θ. Find the magnitude of the acceleration of the two objects and the tension in the cord.

Question

A ball of mass [latex] m_{1} ext { and a block of mass } m_{2} [/latex] are attached by a lightweight cord that passes over a frictionless pulley of negligible mass, as in Figure 5.15a. The block lies on a frictionless incline of angle θ. Find the magnitude of the acceleration of the two objects and the tension in the cord.

Accepted Answer

Conceptualize the motion in Figure 5.15. If [latex] m_{2} [/latex] moves down the incline, [latex] m_{1} [/latex] moves upward. Because the objects are connected by a cord (which we assume does not stretch), their accelerations have the same magnitude. We can identify forces on each of the two objects and we are looking for an acceleration, so we categorize this as a Newton’s second-law problem. To analyze the problem, consider the free-body diagrams shown in Figures 5.15b and 5.15c. Applying Newton’s second law in component form to the ball, choosing the upward direction as positive, yields

[latex] ext { (1) } \quad \sum F_{x}=0 [/latex].

[latex] ext { (2) } \quad \sum F_{y}=T-m_{1} g=m_{1} a_{y}=m_{1} a [/latex].

Note that in order for the ball to accelerate upward, it is necessary that [latex] T>m_{1} g . [/latex] we replaced [latex] a_{y} [/latex] with a because the acceleration has only a y component.
For the block it is convenient to choose the positive x′ axis along the incline, as in Figure 5.15c. For consistency with our choice for the ball, we choose the positive direction to be down the incline. Applying Newton’s second law in component form to the block gives

[latex] ext { (3) } \quad \sum F_{x^{\prime}}=m_{2} g \sin heta-T=m_{2} a_{x^{\prime}}=m_{2} a [/latex].

[latex] ext { (4) } \quad \sum F_{y^{\prime}}=n-m_{2} g \cos heta=0 [/latex].

In (3) we replaced [latex] a_{x^{\prime}} [/latex] with a because the two objects have accelerations of equal magnitude a. Equations (1) and (4) provide no information regarding the acceleration. However, if we solve (2) for T and then substitute this value for T into (3) and solve for a, we obtain

(5) [latex] a=\frac{m_{2} g \sin heta-m_{1} g}{m_{1}+m_{2}} [/latex].

When this expression for a is substituted into (2), we find

(6) [latex] T=\frac{m_{1} m_{2} g(\sin heta+1)}{m_{1}+m_{2}} [/latex].

To finalize the problem, note that the block accelerates down the incline only if [latex] m_{2} \sin heta>m_{1} ext {. If } m_{1}>m_{2} \sin heta ext {, } [/latex] then the acceleration is up the incline for the block and downward for the ball. Also note that the result for the acceleration (5) can be interpreted as the magnitude of the net force acting on the system divided by the total mass of the system; this is consistent with Newton’s second law.

Question 5.10: A ball of mass m1 and a block of mass m2 are attached by a l...

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