Friction on an incline

1. The problem statement, all variables and given/known data

A block is at rest on the incline shown in the figure. The coefficients of static and kinetic friction are µs = 0.67 and µk = 0.57, respectively. The acceleration of gravity is 9.8 m/s^2. What is the frictional force acting on the 42 kg mass? Answer in units of N.

2. Relevant equations

F=MA
f=µn

3. The attempt at a solution

Ok, I thought I understood this one, but when I entered it in the answer was wrong. Here is what I got. Thanks in advance for the help.

MG = 9.8 * 42 = 411.6N

Weight of block in Y direction = 411.6cos31 = 352.810061N

Force down the incline = 411.6sin31 = 211.9896716N

Force down the incline minus friction =
411.6sin31 - µ411.6cos31 = 0
µ411.6cos31 = 411.6sin31
µ = 411.6sin31/411.6cos31
µ = 0.6008606189 N

I entered in 0.6008606189 and it was wrong. Where did I go wrong?
 
136
0
Can you post your FBD? It seems you have missed some forces, and are trying to solve for mu, which is already given.
 
Attached a copy of the actual problem and my work I did on it. Hope it's readable. I just kind of scanned it and pasted it together.
 

Attachments

I was thinking that solving for µ would show the current amount of friction that was being placed on the object at said angle. I thought the 0.67 static friction was the friction from the maximum angle that the incline could be before it slid (inverse tan of 0.67). Where is my thinking off at?
 
136
0
The coefficient of static friction is, technically, independent of the angle. But, yes, raising the incline until the block is how the static friction coefficient is measured/calculated. Since both coefficients are given in the problem, you don't have to try to calculated them. And, since the block is at rest, you can ignore the coefficient of kinetic friction (0.57).


Looking at your diagram, you've started it off right, but all you really need to to now is remember the definition of frictional force - which is really just the second equation you posted in your original post. Also, it's the part you crossed out on your paper. So, you've actually done it. That is the maximum total force (coming from the weight of the block and any potential external force - which isn't present for this problem) that can be applied down the plane before it starts moving.

But, there's two more steps to complete the problem. 1) Compare the two values you have for the forces along the plane. What do you notice? 2) Remember how the block is at rest? What does that mean for the sum of the forces? So, since there's no added external force trying to push the block down the incline, how much frictional force is really holding the block in place?
 
So, if I understand what your saying:

f=µn
f=0.67(411.6cos31)=236.3827408

So the total forces should be the force down the plane minus the y forces * µ:

mgsinΘ-µmgcosΘ = 0 (like i tried earlier but putting the 0.67 in for µ?)
411.6sin31-0.67(411.6cos31) = 0

which equals -24.39306922

Is that right?
 
136
0
Sort of, but I think I gave you too much to think about before. Yes, the maximum available frictional force is 236.38N. Since that force is stronger than the force pulling the block down the incline - which you correctly identified as 411.6N*sin31 (211.99N) - it is confirmed that the block, does in fact, remain at rest. So, since the gravity is pulling the block down the plane with 211.99N, how much force is necessary to keep the block in place?

Ideally, that's how it probably should have been looked at in the first place, with calculating the max available frictional force simply to verify that the block remained static.
 
The block is trying to move down the plane with 211.99N, so the frictional force acting on the object would also be 211.99N? Because if the frictional force was more than 211.99N, it would actually be moving up. Am I on the right track with that line of thought?
 
I entered that answer in and it was right. Thanks for all the help. I was a little confused on exactly what it was asking for at first, but I have a much better understanding of it now. Much appreciated.
 
136
0
Glad to be of some assistance. Friction problems can be tedious at times, especially if you have to verify whether or not the block is moving. It also can be tricky if your given extra info - such as the coefficient of kinetic friction.
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top