A block with mass m1 on a plane inclined at angle to the horizontal is connected to a second hanging block with mass m2 by a cord passing over a small, frictionless pulley, as shown. The coefficient of static friction is us
and the coefficient of kinetic friction is uk
2.1
Find the mass m2 for which block m1, moves up the plane at constant speed once it is set in
To find the mass m2 for which block m1 moves up the plane at a constant speed, we will analyze the forces acting on both blocks.
First, let's consider block m1 on the inclined plane. The gravitational force acting on m1 can be decomposed into two components: one along the plane and one perpendicular to the plane. The component along the plane is m1 * g * sin(α), and the component perpendicular to the plane is m1 * g * cos(α).
Next, let's analyze the forces acting on block m2. There are two forces acting on m2: its weight (m2 * g) and the tension in the cord (T). Since the system is in equilibrium, these two forces must be equal in magnitude. Therefore, we have m2 * g = T.
Since the block m1 is moving up the plane at a constant speed, the forces acting on it must be balanced. The force of static friction (fs) between m1 and the inclined plane opposes the component of the gravitational force along the plane (m1 * g * sin(α)). Therefore, we have fs = m1 * g * sin(α).
The maximum force of static friction (fs_max) can be calculated using the formula fs_max = us * N, where us is the coefficient of static friction and N is the normal force acting on m1. The normal force N is equal to m1 * g * cos(α). Therefore, fs_max = us * m1 * g * cos(α).
Since block m1 is moving up the plane at a constant speed, the static friction force is equal to the force exerted by m2, which is T. Therefore, we have fs_max = T.
Combining the equations, we get:
us * m1 * g * cos(α) = m2 * g
Simplifying the equation, we get:
m2 = (us * m1 * cos(α)) / g
This equation gives us the required mass m2 for which block m1 moves up the plane at a constant speed.