Two boxes sliding down a ramp

[ChrisMC]

Homework Statement

Two packages at UPS start sliding down the 20° ramp shown in Figure P8.25. Package A has a mass of 5.5 kg and a *coefficient of friction of 0.20. Package B has a mass of 10 kg and a coefficient of friction of 0.15. How long does it take package A to reach the bottom?

Homework Equations

F=ma
Net force

The Attempt at a Solution

I found the force of the system to be 15.5gsin(20) then subtracted frictional force, but I really have no idea on what I'm doing.

Any help is appreciated

 

[cepheid]

Welcome to PF!

How long does it take package A to reach the bottom?

You can find the time taken to reach the bottom of the ramp from basic kinematics, since you know that the box has a constant acceleration. However, to do this, you need to know the length of the ramp. Is this information given in the problem?

Also, what is the arrangement of the two boxes? Is there any way that one will hit the other? That would change everything...

Anyway, assuming that they don't, the question is, how do you figure out what that constant acceleration is (for package A)?

As you correctly noted, you need to find the acceleration using Newton's second law, which requires computing the net force. The net force in the "perpendicular to the ramp" direction is zero. The net force in the direction parallel to the ramp is a sum of two opposing forces: the component of the weight which points down the ramp, and friction, which points up the ramp, opposing the motion.

I found the force of the system to be 15.5gsin(20) then subtracted frictional force, but I really have no idea on what I'm doing.

If by "force of the system", you mean the component of the package's weight that points down the ramp, then I agree that this force is given by mgsin(20^o). What did you get for the frictional force? Therefore, what is the net force and the resulting acceleration?

 

[ChrisMC]

thanks,

the total length of the ramp is not given,
the frictional force is just coefficient of friction* normal force which would be

.2(15.5gsin(20))

 

[cepheid]

thanks,

the total length of the ramp is not given,

Is the height of the ramp given? From that you can compute the length.

the frictional force is just coefficient of friction* normal force which would be

.2(15.5gsin(20))

In your first post, you claimed that the parallel component of the weight is mgsin(20^o). Now you are claiming that the perpendicular component of the weight is given by that as well. Clearly, both of these statements cannot be true. Double check your trig.

Edit: Also, why are you using the combined mass of both boxes when the problem is clearly only asking about box A? Again, I will ask you, are they touching each other? If so, that changes things.

 

[ChrisMC]

yes they are touching eachother, I used the combined mass because its a system

 

[cepheid]

Which one is in front? If the "faster" (lower friction) box is in front, then they won't remain touching. If the faster one is in the back, then the friction force of the slower one determines the net force on it, and a contact force arises between the boxes to make up the difference.

Do you have enough info now to solve this?

 

[ChrisMC]

the .2 one is in the front, I still dont know what the equation would be

 

[cepheid]

Then they stay connected, since the slower one is in front. You know that the two boxes have the same accelerations, in spite of the fact that they have different frictional forces acting on them. How is this possible? It's possible because there are also contact forces in between the boxes. Box A pushes on Box B, and Box B pushes back on Box A. You know that these two contact forces must be equal to each other by Newton's third law. The fact that these forces are equal, and that the accelerations are equal, gives you the constraints you need to solve this system.

Draw a free body diagram for each box separately, and it'll become clearer what I'm talking about.