Inclined Planes and Two Masses

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Homework Statement


Two blocks connected by a cord passing over a small fictionless pulley rest on an double inclined plane (ie a traingle) with static friction coefficient of 0.5 and a kinetic friction coefficient of 0.4. The mass of block A is 100 kg (sitting at 30 degrees), while the mass of block B is 50 kg (sitting at 53.1 degrees).

A. Which way will the system move when the blocks are released from rest?
B. What is the acceleration of the blocks?


Homework Equations


For A, we need to look at the forces pulling down on the block right? So for mass A is it, mgsinӨ-u(kinetic)mgcosӨ=151N and for mass B is it, mgsinӨ-u(kinetic)mgcosӨ=274N. So mass B must slide down, while mass A must slide up right?

For B, the F=ma equation for mass A is T-mgsinӨ-u(kinetic)mgcosӨ=ma, while for mass B, it is mgsinӨ-u(kinetic)mgcosӨ-T=ma.

So T-151=100a, meaning T is 151+100a. Plugging into the equation for mass B, 391.8-117.7-(151+100a)=50a. So a=0.821 m/s^2. Is this right?
 

Answers and Replies

  • #2
ehild
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Homework Statement


Two blocks connected by a cord passing over a small fictionless pulley rest on an double inclined plane (ie a traingle) with static friction coefficient of 0.5 and a kinetic friction coefficient of 0.4. The mass of block A is 100 kg (sitting at 30 degrees), while the mass of block B is 50 kg (sitting at 53.1 degrees).

A. Which way will the system move when the blocks are released from rest?
B. What is the acceleration of the blocks?


Homework Equations


For A, we need to look at the forces pulling down on the block right? So for mass A is it, mgsinӨ-u(kinetic)mgcosӨ=151N and for mass B is it, mgsinӨ-u(kinetic)mgcosӨ=274N. So mass B must slide down, while mass A must slide up right?

the F=ma equation for mass A is T-mgsinӨ-u(kinetic)mgcosӨ=ma, while for mass B, it is mgsinӨ-u(kinetic)mgcosӨ-T=ma.

So T-151=100a,

Check the red line.


ehild
 
  • #3
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Check the red line.


ehild

So it's

For part B, the F=ma equation for mass A is T-mgsinӨ-u(kinetic)mgcosӨ=ma, while for mass B, it is mgsinӨ-u(kinetic)mgcosӨ-T=ma.

So T-829=100a, meaning T is 829+100a. Plugging into the equation for mass B, 391.8-117.7-(829+100a)=50a. So a=-11.1 m/s^2. Is this right?

But then doesn't that mean that mass A is falling down the incline, instead of mass B?
 
  • #4
ehild
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But then doesn't that mean that mass A is falling down the incline, instead of mass B?

Strange isn't it? You have proved that A can not descend, and now you got that B doesn't move downward. There is a third possibility: neither blocks move. Is it possible? You are given the coefficient of static friction.


ehild
 

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