What Is the Minimum Weight of Block C to Prevent Block A from Sliding?

[farleyknight]

Homework Statement

In Fig. 6-37, blocks A and B have weights of 51 N and 24 N, respectively. (a) Determine the minimum weight of block C to keep A from sliding if μ_s between A and the table is 0.20. (b) Block C suddenly is lifted off A. What is the acceleration of block A if μk between A and the table is 0.10?

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c06/fig06_35.gif

Homework Equations

The Attempt at a Solution

I started to draw some free body diagrams, but it got rather carried away. There's only a handful of forces involved, so instead, I will analyze block B then block A, giving the forces that are affecting them each.

On block B, there are exactly 2 forces. The first force is the force of gravity, call it F_gB, in a downward, negative y-axis, direction. Opposing this force, is the force of tension, call it T, which has equal magnitude but opposite direction. Then we have

F_gB = T

for both magnitudes.

For block A, we have 4 forces. The first force is the T mentioned earlier, caused by the string. It moves in the positive x-direction. We also have the force of gravity, caused by both block A and block C. For that, we'll denote

F_g = m_A * g + m_C * g = m_A * g + W_C

With W_C being the weight of block C. The normal force F_N is just the opposite of the force of F_g, but point up. The force of friction is moving toward the left, and given the coefficient of friction, we have

F_f = mu_k * F_N = mu_k * (m_A*g + W_C)

Since block A refuses to move, we have

F_net,x = 0 = T - F_f

Calculating T and F_f, we have

T = m_B * g = 24*9.8 = 235.2

F_f = 0.2*(51*9.8 + W_C) = 99.96 + 0.2*W_C

T - F_f = 235.2 - (99.96 + 0.2*W_C)

W_C = 135.24 / 0.2 = 676.2 N

But this is not correct. I've come up with the same answer twice. My feeling is that messed up on the force of gravity on block A, or the normal force on A.. I'm not completely sure.

 

[206PiruBlood]

I would just consider that block A is being pulled to the right by block B; in order for the system to remain static the frictional force must counter this.

 

[farleyknight]

I would just consider that block A is being pulled to the right by block B; in order for the system to remain static the frictional force must counter this.

That's pretty much what I have done, except that my solution for block C is not correct.. I'm not sure what I've done wrong in my calculations, but I suspect that the normal force is not the combined gravities of A and C..

 

[206PiruBlood]

I believe your error is where you calculated friction. You know that the AC block system needs a frictional force of 24N to counter the 24N block B. What then is the net required normal force to achieve 24N of friction with 0.2 coefficient?

 

[bleedblue1234]

In Fig. 6-37, blocks A and B have weights of 51 N and 24 N, respectively. (a) Determine the minimum weight of block C to keep A from sliding if μ_s between A and the table is 0.20. (b) Block C suddenly is lifted off A. What is the acceleration of block A if μk between A and the table is 0.10?

so f = 51N*.20

f=10.2N

so 24N-10.2N = 13.8N to the right (unbalanced force)

so

13.8N = C * .20

so C = 69N

and F = ma

so 13.8N = 51N * a

so a = .271 m/s/s

 

[206PiruBlood]

In Fig. 6-37, blocks A and B have weights of 51 N and 24 N, respectively. (a) Determine the minimum weight of block C to keep A from sliding if μ_s between A and the table is 0.20. (b) Block C suddenly is lifted off A. What is the acceleration of block A if μk between A and the table is 0.10?

so f = 51N*.20

f=10.2N

so 24N-10.2N = 13.8N to the right (unbalanced force)

so

13.8N = C * .20

so C = 69N

and F = ma

so 13.8N = 51N * a

so a = .271 m/s/s

Your second answer is wrong(for multiple reasons). Also, you should't just do the problem for him!

 

[hotvette]

In Fig. 6-37, blocks A and B have weights of 51 N and 24 N,...

Your approach is valid. The mistake is that you treated the blocks as masses instead of weights.

 

[farleyknight]

Your approach is valid. The mistake is that you treated the blocks as masses instead of weights.

Thank you. I managed to catch that and answer the first question. However, the second question still isn't working out.

If block C is lifted, then the new normal force on a (F_N) simply becomes it's own weight. And then the frictional force is just F_f = mu_k * F_N = 0.1 * 51 = 5.1. Counteracting that against the tension in the cord, we would have

F_net,x = m*a = T - F_f = 24 - 5.1

To find the mass of block A, we simply set 51 = m*g = m*(9.8). So the mass of block A = 5.2. And then

F_net,x = m*a = 5.2*a = 18.9.

So then a = 3.63. But that value isn't working out :(

I have a feeling that the normal force isn't going to be F_g on block A, since the act of lifting might have some kind of upward effect. But I have no idea.

 

[ideasrule]

F_net,x = m*a = T - F_f = 24 - 5.1

Tension isn't equal to 24 N because the "B" block is now accelerating. If you write out Newton's second law for the block, you'd get mg-T=ma, T=m(g-a).

 

[farleyknight]

Tension isn't equal to 24 N because the "B" block is now accelerating. If you write out Newton's second law for the block, you'd get mg-T=ma, T=m(g-a).

Thank you for pointing that out. I managed to fix it :)