Two masses on an incline help

[Crazdfanatic]

Homework Statement

Two boxes, with masses 12.5 kg and 25.5 kg, are connected by a taut string on a surface that is inclined at 25 degrees with respect to the horizontal. If the coefficients of kinetic friction between the boxes and the table are 0.100 and 0.150 respectively and the 25.5 kg box is placed below the 12.5 kg box on the incline, what is the magnitude of the acceleration of the 12.5 kg box just after they have begun to move down the incline?

The answer we are looking for is supposed to be 4.14 m/s^2

Homework Equations

F = mg*sin(theta) - umg*cos(theta)

The Attempt at a Solution

I first assumed that since the coefficient of friction of the the heavier block was higher, it would have no bearing on the lighter block's acceleration. This yielded the answer 3.26 m/s^2, which was wrong.

I then tried taking the forces produced by the blocks separately(12.5*9.81*sin25 - .100*12.5*9.81*cos25, and then the same but with the data for the heavier block), and added them together to get the overall force, and then divided by the sum of their masses yielding 2.96 m/s^2 as my answer, but to no avail.

Lastly, I tried adding the separate forces again, but this time diving by just the mass of the lighter weight. Once again, wrong.

I seem to of come to a roadblock in my thoughts, so any help would be appreciated.

If something I've stated is unclear, I apologize, just tell me and I'll try to explain.

 

[anathema]

Hello! It seems like you are on the right track with your attempts. However, there are a few things to consider in this problem.

First, when calculating the force of friction, it is important to remember that the force of friction is always in the opposite direction of motion. In this case, the boxes are moving down the incline, so the force of friction will be pointing up the incline.

Secondly, the coefficient of friction is a measure of how difficult it is for the two surfaces to slide against each other, so it will have an effect on the acceleration of both boxes, not just the heavier one.

With these factors in mind, here is an approach you can try:

1. Draw a free body diagram for each box. This will help you visualize all the forces acting on each box.

2. Write out the equations for the sum of forces in the x and y directions for each box. Remember to consider the force of friction and the angle of the incline.

3. Use the equations to solve for the acceleration of each box.

4. Since the boxes are connected by a string, they will have the same acceleration. Use this to find the overall acceleration of the system.

I hope this helps! Let me know if you have any further questions or if you need any clarification on the steps. Keep up the good work!

 

[Moetasim]

I would approach this problem by first drawing a free body diagram for each box, showing all the forces acting on them. This would include the gravitational force, the normal force, and the frictional force. I would then use Newton's Second Law, F=ma, to set up equations for each box and solve for the acceleration.

In this case, we have two equations:

For the 12.5 kg box:
F = 12.5 kg * a
F = mg*sin(theta) - umg*cos(theta)

For the 25.5 kg box:
F = 25.5 kg * a
F = mg*sin(theta) - umg*cos(theta)

We can then solve for the acceleration by setting the two equations equal to each other:
12.5 kg * a = 25.5 kg * a
mg*sin(theta) - umg*cos(theta) = mg*sin(theta) - umg*cos(theta)
Canceling out the common terms, we are left with:
a = 0

This means that the acceleration of both boxes is 0, and they will not move down the incline. This could be due to the fact that the frictional force is greater than the gravitational force in this case.

To get an acceleration of 4.14 m/s^2, we would need to adjust the coefficients of friction or the angle of the incline. I would also suggest checking your calculations and units to ensure accuracy.