Inclined Planes and Two Masses

[dododo121]

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?

 

[ehild]

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

 

[dododo121]

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]

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

 

[Nickie]

Your equations and calculations for A and B are correct. Based on the forces and angles involved, block B will slide down while block A will slide up. The acceleration of the blocks will be 0.821 m/s^2, with block B accelerating down the incline faster than block A accelerates up the incline. This is due to the difference in mass and the angle of the incline for each block. It is important to consider both static and kinetic friction coefficients in order to accurately calculate the acceleration of the system.