# Inclined Planes and Two Masses

## 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?

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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

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?

ehild
Homework Helper
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