# Homework Help: Newtons Law and Friction Question (along with a contact force question)

1. Jun 21, 2010

### iurod

1. The problem statement, all variables and given/known data
Two crates, of mass 75 kg and 110 kg are in contact and at rest on a horizontal surface. A 620N force is exerted on the 75 kg crate. If the coefficient of kinetic friction is 0.15 calculate:
a) the acceleration of the system
b)the force that each crate exerts on the other

* I have attached a word document with the picture my book gave me, hoping that it helps.

2. Relevant equations
f=ma
Friction= (coefficient of kinetic friction)(mg)

3. The attempt at a solution

a) the acceleration of the system

for the Free Body Diagram I chose to make one big block with mass= 185kg

f=ma
620N - Friction = (185 kg)a
Friction= (0.15)(185x9.8) = 271.95

620 - 271.95 = (185 kg)a
a= 1.9 m/s2

b. the force that each crate exerts on the other

This is where I'm having trouble, I could find the formula nor did our professor give us the formula for Contact force. So I tried to think it out and got this

Contact Force = Force - Friction - ma

taking the 75kg box first:
620 - Friction -(75 kg)(1.9 m/s2)
Friction = (0.15)(75x9.8) = 110.25

Contact force = 620 - 110.25 - (75)(1.9)
Contact force = 367.25 I rounded to 3.7x102N

Now for the 110 kg box:
620 - Friction - (110)(1.9)
Friction = (0.15)(110x9.8) = 161.7

Contact force = 620 - 161.7 - (110)(1.9)
Contact force = 249.3 I rounded to 2.5x102

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

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### problem 48 figure.doc
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73.5 KB
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2. Jun 22, 2010

### PhanthomJay

The contact force of the 75 kg block on the 110 kg block must be equal and opposite to the contact force of the 110 kg block onthe 75 kg block, which you calculated correctly as having a magnitude of 370 N. This follows from Newton's 3rd law. So you made an error when you looked at the free body diagram of the 110 kg block by introducing a force on that block that does not exist. Which force doesn't belong? Also, in which direction does the contact force act in this case?

3. Jun 22, 2010

### iurod

I follow. I think I got confused when doing the problem. Is this correct?

So for Question b:
I got the Contact force as 370N of the 75kg block on the 110kg block and -370N of the 110kg block on the 75kg block.

Now for part c:
If I go to the other side of the table and apply a 620N Force to the 110kg.
The contact force on the 110kg block on the 75kg block is 250N and the 75kg block on the 110kg block is -250kg.

4. Jun 22, 2010

### PhanthomJay

Your magnitudes are correct, but i couldn't open your figure to see what direction you mean by the minus sign. The contact force in any case is a force that pushes on the other block.

5. Jun 22, 2010

### iurod

The picture just has a table with the two blocks on top of it with an arrow pointing to the blocks, indicating the 620N force.

I put the minus sign because Newtons Third Law "for every action there is an equal but opposite reaction" So if the 75kg block has a contact force of 320 on the 110kg block. Should the 110kg block have an equal but opposite reaction to the 75kg block? making it -320N?

Same thing goes for the second case...

Thanks again for helping me with this.

6. Jun 22, 2010

### PhanthomJay

Yes, you are right about Newton 3, so the contact forces on each respective block point in opposite directions. For the first case, with the 620 N force applied to the 75 kg block, when you look at the contact force of the 110 kg block on the 75 kg block, it should point in the opposite direction of the 620 N force. Is that what you show?

7. Jun 22, 2010

### iurod

hmmm. what I did was first find the contact force which I set up like this:

Contact force = Force - Ffriction - ma
Contact force = 620N - 110.25N - (75x9.8)
Contact force = 367.25 but I rounded to +370N

the + is the contact force exerted on the 110kg box by the 75kg box. So for the contact force exerted on the 75kg box by the 110kg its the same magnitude but with opposite sign; -320N

I did the same thing for the second case, but this time the initial 620N force is applied to the 110kg.