Question

Problem Set 4 1. Block and Pulley System Block 1 m1 Pulley 2 m2 A block of mass m₁ is at rest on an inclined plane that makes an angle with the horizontal. The coefficient of static friction between the block and the incline surface is . A massless, inextensible string is attached to one end of the block, passes over a fixed pulley, pulley 1, around a second freely suspended pulley, pulley 2, and is finally attached to a fixed support. The left hand part of the string is parallel to the surface of the inclined plane. The sections of the string coming off the suspended pulley are vertical. The pulleys are massless, but a second block of mass m₂ is hung from the suspended pulley. Gravity acts downward. (a) What is m2,min, the minimum value of m2 for which block 1 just barely slides up the incline? Express your answers in terms of some or all of the variables m₁, 0, s and g. (b) What is m2,max, the maximum value of m₂ for which block 1 just barely slides down the incline? Express your answers in terms of some or all of the variables m₁, 0, μs and g. (c) Now assume that the block on the incline plane is sliding upward. The coefficient of kinetic friction between the block and the incline surface is k. Find the magnitude of the acceleration of the block on the inclined plane, a1. Express your answers in terms of some or all of the variables m₁, m2, 0, k and the acceleration of gravity g.​

Answer 1

(a) To find the minimum value of m₂ for which block 1 just barely slides up the incline, we need to consider the forces acting on block 1. These forces are the gravitational force (mg) acting vertically downwards and the force of friction (Ff) acting parallel to the incline and in the opposite direction to the motion.

Since the block is just about to slide up the incline, the force of friction is at its maximum value, which is equal to the product of the coefficient of static friction and the normal force. The normal force is equal to the component of the weight of block 1 that is perpendicular to the incline, which is mg cosθ.

Thus, we have Ff = μs mg cosθ.

The tension in the string is equal to the weight of block 2, which is m₂g. Since the pulleys are massless, the tension is the same throughout the string.

Using Newton's second law, we can write the equation of motion for block 1 as:

mg sinθ - Ff = m₁a

where a is the acceleration of block 1 up the incline.

Substituting the expression for Ff, we get:

mg sinθ - μs mg cosθ = m₁a

Rearranging, we get:

m₂ = (m₁ + μs m₁ tanθ) / (μs - sinθ)

This is the minimum value of m₂ for which block 1 just barely slides up the incline.

(b) To find the maximum value of m₂ for which block 1 just barely slides down the incline, we need to consider the forces acting on block 1. These forces are the gravitational force (mg) acting vertically downwards and the force of friction (Ff) acting parallel to the incline and in the direction of motion.

Since the block is just about to slide down the incline, the force of friction is at its maximum value, which is equal to the product of the coefficient of static friction and the normal force. The normal force is equal to the component of the weight of block 1 that is perpendicular to the incline, which is mg cosθ.

Thus, we have Ff = μs mg cosθ.

The tension in the string is equal to the weight of block 2, which is m₂g. Since the pulleys are massless, the tension is the same throughout the string.

Using Newton's second law, we can write the equation of motion for block 1 as:

mg sinθ + Ff = m₁a

where a is the acceleration of block 1 down the incline.

Substituting the expression for Ff, we get:

mg sinθ + μs mg cosθ = m₁a

Rearranging, we get:

m₂ = (m₁ - μs m₁ tanθ) / μs

This is the maximum value of m₂ for which block 1 just barely slides down the incline.

(c) When the block on the incline plane is sliding upward, the force of friction is kinetic and is given by:

Ff = k mg cosθ

where k is the coefficient of kinetic friction.

The equation of motion for block 1 is:

mg sinθ - Ff = m₁a₁

Substituting the expression for Ff, we get:

mg sinθ - k mg cosθ = m₁a₁

Rearranging, we get:

a₁ = g (sinθ - k cosθ)

This is the magnitude of the acceleration of block 1 up the incline when it is sliding upward.