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## GR0877 Problem 58

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (A) The only force that provides acceleration $a$ for the block B is the friction force due to the block A. Therefore, by Newton’s second law, this force is equal to $m_B a = 10$ kg $\cdot$ 2 $\text{m/}\text{s}^2= 20$ N.

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Categories: 0877 Tags: , , , ,
1. 22.10.2012 at 19:55

Wouldn’t this only hold true only if the entire force were being applied to A alone? Only then the frictional force due to A will be responsible for the acceleration of B, right?

2. 29.03.2014 at 20:35

same thoughts here. Where does it say in the question that we can assume the only force on the top block is friction?

3. 04.10.2014 at 22:47

Why do I get the wrong answer if I calculate the frictional force by multiplying the normal force by the coefficient of static friction?

• 29.08.2015 at 01:27

Because fs = mu_s*N is the maximum value for static friction that you’d have to overcome to get B moving off A. The value of static friction can be lower than this maximum value. In this problem it is lower since the force applied isnt very much compared to mu_s*N.

4. 19.09.2015 at 06:09

There is clearly an external force acting in this problem with no indication of which block it is acting on (if not both). Unless I’m missing something – terrible problem!

5. 22.10.2015 at 08:32

Yeah, I concur, this is a terrible problem

6. 29.08.2016 at 09:42

This is a bad problem, but nevertheless one can at least go about answering this faulty question by seeing that if you one assumes the acceleration applies to BOTH blocks A and B, the answer (80 N) isn’t even one of the answer choices.