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In the figure, blocks A and B have weights of 45 N and 23 N, respectively. (a) Determine the minimum weight of block C to keep A from sliding if μs between A and the table is 0.21. (b) Block C suddenly is lifted off A. What is the acceleration of block A if $\mu_k$ between A and the table is 0.14?

enter image description here

I found the weight of block C to be $64.5 N$ , and I was told that was correct.

For part b, here is my work.

$\mu_k = 0.14$

$F_k = \mu_k F_N$ $\text{ }$ $F_k = \text{frictional force due to } \mu_k$

$F_k = 0.14 * 45N = 6.3N$

$F_{net} = F_B - F_k$

$F_{net} = 23N - 6.3N = 16.7N$

$F = ma$

$16.7 = \frac{45N}{9.81 {m\over s^2}}*a$

$a = 3.64 {m\over s^2}$

However, I am told that this answer is wrong. Did I make a mistake anywhere? Any help would be appreciated!

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  • $\begingroup$ The figure link is not working. Try putting the image to imgur.com $\endgroup$ – ja72 May 29 '13 at 17:48
  • $\begingroup$ Basically, all the figure shows is a block A with a block C sitting on top of it, then a rope attached to block A going over a frictionless, massless pulley which is attached to block B, and block B is suspended in the air. The same picture can be found at this link as well: has.vcu.edu/new-phy/WSG/CH6%20images/prob25.pdf $\endgroup$ – Sammy May 29 '13 at 17:59
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$F_{net}=F_B-F_{k}$ is wong.

It should be $F_{net}=F_{T}-F_{k}$ , $F_{t}$ is the tension force.

You have to write equations of motion for A,B to get the tension force or acceleration.

Another way to solve these problems would be to consider:

$$\text{acceleration}(a)=\dfrac{\text{net force on the system}}{\text{net inertia of system}}^*$$

In your case net force is what you write $F_{net}$ and net inertia is total mass of the blocks.

*This may not work for complicated systems moving with different accelerations

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