# Newton's Laws and Friction Applied to Blocks

1. Oct 27, 2006

### BSCS

Given: 2 wedge shaped blocks arranged as a rectangle.

The bottom block is on a frictionless horizontal surface.

Horizontal force Q is applied to the vertical side of the bottom block.

Find Q max, the maximum value of Q where there is no slipping, in terms of us, m, and theta.

Masses: m and m
us is the coefficient of static friction between the blocks.

Code (Text):

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|  /|
| / |<---Q
|/__|

Part 2: If the force is applied to the other block horizontally, what is Q max?

Code (Text):
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|  /|
Q-->| / |
|/__|

For the first part, I applied newton's laws to the top block and got the correct answer.

For the second part, If I apply newton's laws to the top block again, I get the correct (same) answer; BUT only if I ignore Q in my equations for F=ma.

From my reasoning, the top block will be the one slipping under both circumstances, so that's the block that I addressed...

Is this strategy correct? If so, why can I ignore Q? If not, how do I approach this?

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Here is my thinking, please tell me if this is correct:

Since we need Fmax, we can find the maximum acceleration such that there is about to be slipping. Because there is no slipping at this point, the accelerations of each block, as well as of the system, are all equal. So, by figuring out the maximum acceleration of the top block, which is the one that will slip upward on the other block, we can use this to solve the problem.

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Also, does anyone know where I can find more problems like this?

Last edited: Oct 28, 2006
2. Oct 27, 2006

### PhanthomJay

You noted that you applied newton's laws to the top block, but you must also apply newton's laws to the system of blocks, which of course you must have done to get the correct answer to part (a). Yes, the accelerations of each block and trhe system of blocks is the same.
In part (b), you must again take the FBD of the system and top block, and you will arrive at the same answer for Q, but i don't see how you could have ignored Q in ther FBD of the top block , since it is there, and how you could have arrived at the same (correct) result. In both parts a and b, for the system, Q =2ma, that is, ma =Q/2. In part a, without getting into the math, in the horizontal direction you have, for the top block
N(costheta +usintheta) = ma = Q/2
and in part b, isolating again the top block and lookiong in the horizontal direction,
Q - N(costheta + usintheta) =ma =Q/2, from which
Q-Q/2 = N(costheta + usintheta) = Q/2
which is the same result as a, and the correct result(after plugging in the value of N determined from the vertical direction equilibrium equations).

3. Oct 28, 2006

### BSCS

Thanks for the help. I tried the problem over again using ma=Q/2 and was able to easily solve both (a) and (b). FYI The correct answer is listed as having the cos and sin reversed from what you said...

I'm a bit concerned though...

I was shown how to do part (a) as follows:

Find amax for the top block (using sum(F)x=ma for the top block in the horizontal, and sum(F)y=0, solving for a) and then plug that into F=ma for the system. This worked out to Q=2mamax

This worked fine for part (a), but for part (b) I was getting:

amax= Q - g(u*cos(theta)-sin(theta))/(cos(theta)-u*sin(theta))

F = ma for the system:

Q = mamax

Q = 2m * (Q - g(u*cos(theta)-sin(theta))/(cos(theta)-u*sin(theta)))

Q = 2mQ - 2mg(u*cos(theta)-sin(theta))/(cos(theta)-u*sin(theta))

2mQ - Q = 2mg(u*cos(theta)-sin(theta))/(cos(theta)-u*sin(theta))

2mQ - Q = 2mg(u*cos(theta)-sin(theta))/(cos(theta)-u*sin(theta))

Q(2m - 1) = 2mg(u*cos(theta)-sin(theta))/(cos(theta)-u*sin(theta))

Q = 2mg(u*cos(theta)-sin(theta))/( (cos(theta)-u*sin(theta))(2m - 1))

Messier algebra, and wrong... Did I just make a mistake in the algebra?

Last edited: Oct 28, 2006
4. Oct 28, 2006

### PhanthomJay

I wrote my answer for Q in terms of N, where N is the normal force between the 2 wedges perpendicular to the incline, and not to be confused with the normal force actiing on the botttom wedge from the floor. You've got to solve for N using vertical direction equations. Now i'm so confused with plus and minus signs and haven't taken the time to verify result, however, I note you have a major error somewhere in this messy algebra where you state correctly that Q=2ma_max by looking at the entire system, but then I think when isolating the top wedge in part b, uou state Q=ma_max, which is incorrect, since there are other horizontal forces involved in that motion equation in the x direction.

5. Oct 28, 2006

### OlderDan

Annotations are in the quote in blue.

Last edited: Oct 28, 2006
6. Oct 28, 2006

### BSCS

Thanks! I got it. Boy, do I feel stupid. I think it looks ok, now, right? (this isn't homework, btw, it's just extra

practice) I made 2 algebraic errors:

1) when I pulled out the N, I didn't change the - to a +
2) when I divided by m, I didn't divide Q by m

I found them while typing up my work step by step to post... Maybe there's a lesson there for me.

Part (b): Taking +x to the right, and +y up.

Top Block:

$$\sum F_{x} = Q - f_{k}cos\theta - Nsin\theta = ma_{max}$$

$$Q - \mu Ncos\theta - Nsin\theta = ma_{max}$$

$$Q - N(\mu cos\theta + sin\theta) = ma_{max}$$

$$\sum F_{y} = Ncos\theta - f_{k}sin\theta - mg = 0$$

$$Ncos\theta - \mu Nsin\theta = mg$$

$$N(cos\theta - \mu sin\theta ) = mg$$

$$N = mg / (cos\theta - \mu sin\theta )$$

Substituting N:

$$Q - mg (\mu cos\theta + sin\theta) / (cos\theta - \mu sin\theta ) = ma_{max}$$

$$a_{max} = Q/m - g (\mu cos\theta + sin\theta) / (cos\theta - \mu sin\theta )$$

System:

F = ma

$$Q = 2m (Q/m - g (\mu cos\theta + sin\theta) / (cos\theta - \mu sin\theta ))$$

$$Q = 2Q - 2mg (\mu cos\theta + sin\theta) / (cos\theta - \mu sin\theta ))$$

$$Q = 2mg (\mu cos\theta + sin\theta) / (cos\theta - \mu sin\theta ))$$

7. Oct 28, 2006

### PhanthomJay

It looks good to me. But I don't understand why you thought this was an easy problem. It was not. Especially in part b, where you had to correctly identify, in the FBD of the top wedge , the direction and angle of the 4 forces acting on it, while noting also that there was no 5th force pushing up from the floor. Moderately difficult, i would say.

8. Oct 29, 2006

### BSCS

I didn't mean to give the impression that I thought it was easy. I felt stupid because I made algebraic mistakes that I shouldn't have made. After spending so much time on the problem, I guess I was getting used to the "situation" in the problem... Thanks again.

Last edited: Oct 29, 2006