1.1 A ball of mass m1 and a block of mass m2 are attached by a light weight cord that passes over a frictionless pulley of negligible mass as in Figure 1. The block lies on a frictionless incline of angle θ. Determine the magnitude of the acceleration of the two objects and the tension in the cord. Figure 1
Question
1.1 A ball of mass m1 and a block of mass m2 are attached by a light weight cord that passes over a frictionless pulley of negligible mass as in Figure 1. The block lies on a frictionless incline of angle θ. Determine the magnitude of the acceleration of the two objects and the tension in the cord.
Figure 1
Solution
To solve this problem, we need to apply Newton's second law (F = ma) to both the ball and the block.
Step 1: Identify the forces acting on each object.
For the ball, the only force acting on it is its weight (m1*g), acting downwards.
For the block, there are two forces acting on it: its weight (m2g), acting downwards, and the normal force from the incline. However, because the block is on an incline, we need to consider the components of the weight that are parallel and perpendicular to the incline. The component of the weight parallel to the incline is m2gsin(θ), and the component of the weight perpendicular to the incline is m2g*cos(θ).
Step 2: Apply Newton's second law to each object.
For the ball, we have m1a = m1g - T, where T is the tension in the cord.
For the block, we have m2a = T - m2g*sin(θ).
Step 3: Solve the two equations simultaneously to find a and T.
Adding the two equations gives m1a + m2a = m1g - m2gsin(θ), which simplifies to (m1 + m2)a = m1g - m2gsin(θ). Solving for a gives a = (m1g - m2gsin(θ)) / (m1 + m2).
Substituting this expression for a into the equation for the ball gives T = m1g - m1a = m1g - m1(m1g - m2gsin(θ)) / (m1 + m2) = m1m2g(1 - sin(θ)) / (m1 + m2).
So the acceleration of the two objects is (m1g - m2gsin(θ)) / (m1 + m2), and the tension in the cord is m1m2g(1 - sin(θ)) / (m1 + m2).
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