2 Block Atwood Machine Inclined Plane Problem

Question

A block of mass mi is placed on an inclined plane with a slope angle and connected to a second hanging block m2 by a light string passing over a light frictionless pulley. The coefficients of static and kinetic friction between block 1 and the inclined plane are us and uk.

a. Find the maximum value of m2 for which block 1 remains at rest on the

inclined surface.

b. Find the value of mass m2 for which block 1 moves up the inclined at a constant speed.

c. Now suppose that m2 is large enough that block 1 accelerates up the inclined. Find the acceleration of the system of two blocks.

(specifically I need help with part A and part C)

Answer 1

Given

Block 1 Mass: M1

Block 2 Mass: M2

Angle of Incline: θ

Solution:

A) Max Value of M2 for which block 1 remains at rest.

M2 must be greater then M1 and M1 must experience impending motion for it to remain at rest and meet the Max value of M2 criteria.

Impending Motion: F = fs = μ_s

Draw Free Body Diagrams and apply equilibrium equations (Sum force X and Y equal to Zero)

FBD of Block 1

Normal force perpendicular to the incline

Friction force along the incline (Points to the left along the incline, since block 1 is impending up the incline)

Force of Weight is vertical down

Force of tension is along the incline (Points to the right along the incline)

Sum force Y equal to Zero.

N - m1*g*cosθ = 0

N = m1*g*cosθ (Equation 1)

Sum force X equal to Zero

T - μ_s*N - m1*g*sinθ = 0

T = m1*g * (μ_s*cosθ+sinθ) (Equation 2)

Now, draw Free body diagram of block 2

Tension points upward

Force of Weight points down

Apply sum force Y equal to zero

T - m2*g = 0 (Equation 3)

Sub in Equation 2 into 3.

m1*g * (μ_s*cosθ+sinθ) - m2*g = 0

Solve for m2

Part A Answers

m2 = m1*(μ_s*cosθ+sinθ)

b) Block 1 moves up at constant speed. Find m2

Constant speed means acceleration on blocks is zero.

Motion means the friction is kinetic, therefore fk = μ_k

Same force directions as part a...therefore we simply make the adjustment on friction constant

m2 = m1*(μk*cosθ+sinθ)

c) Acceleration on blocks occur. M1 goes up the incline. M2 goes down.

FBD of Block 1

Normal force perpendicular to the incline

Friction force along the incline (Points to the left along the incline, since block 1 is moving up the incline)

Force of Weight is vertical down

Force of tension is along the incline (Points to the right along the incline)

Sum force Y equal to 0

N - m1*g*cosθ = 0

N = m1*g*cosθ (Equation 1)

Sum force X equal to m1a

T - μ_s*N - m1*g*sinθ = m1a

T = m1*g * (μ_k*cosθ+sinθ)+m1a (Equation 2)

Now, draw Free body diagram of block 2

Tension points upward

Force of Weight points down

Apply sum force Y equal to m2a

T - m2*g = m2a (Equation 3)

Sub in Equation 2 into 3.

m1*g * (μ_k*cosθ+sinθ)+m1a - m2*g = m2a

m1*g * (μ_k*cosθ+sinθ) - m2*g =a(m2-m1)

g * (m1*μ_k*cosθ+m1*sinθ-m2)=a(m2-m1)

a =( g * (m1*μ_k*cosθ+m1*sinθ-m2))/(m2-m1)