Question

rope Given that two boxes, 10 kg and 15 kg respectively, are connected with an inelastic according to the diagram show below. A force of 500 N is applied. The boxes move with an acceleration of 2 ms. One third of the total frictional force is acting on the 10 kg block and two thirds on the 15 kg block. Calculate: (a) the magnitude and direction of the frictional force present. (b) the magnitude of the tension in the rope at T. a = 2 ms? 500 N 10 kg T 15 kg​

Answer 1

Answer:

If a and b are integers, we say a divides b if there exists an integer k so that b=ak. Thus:

b=ak

a=bj

a=akj

kj=1

Since a divides b and b divides a, we know that k,j must both be integer, which means k and j must both equal 1 or -1. This means:

a=b

This proves that a=b when we assume that a,b are both positive integers.

For the second part of the question, I am asked if I can still say a=b if the assumption about a, b is relaxed so that a,b belong to the integers (not the positive integers.) I think The answer is no, I cannot say that a=b for every case when we allow a,b to also be negative integers.

Here is my reasoning (I use the same idea that a=bj and b=ak here)

+/−a=akj

+/−1=kj

So |a|=|b| but not a=b

I'm very new to proofs so any suggestions or thoughts are greatly appreciated.

Answer 2

Answer:

As long as the applied force has a magnitude smaller than μsN, the force of static friction fs has the same magnitude as the applied force, but points in the opposite direction. The magnitude of the force of kinetic friction acting on an object is fk = μkN, where μk is the coefficient of kinetic friction.