Q5. Two trunks sit side by side on the floor. The larger...

Q5. Two trunks sit side by side on the floor. The larger trunk (52 kg) is to the left of the smaller trunk (34 kg). A person pushes on the larger trunk horizontally toward the right. The coefficient of static friction between the trunks and the floor is 0.35. (a) Determine the magnitude of the maximum force the person can exert without moving either trunk. (b) Calculate the force the larger trunk exerts on the smaller trunk. (c) Would either answer change if the person pushed in the opposite direction on the smaller trunk? Explain your reasoning. 4.0 kg pulley 1.8 kg

Transcript

00:01 Trunk a of mass m1 equal to 52 kg is on the left side and trunk b of mass m2 equal to 34 kg is on the right side.

00:16 They are lying on a floor and the coefficient of static friction between the floor and the trunks is mu equal to 0 .35.

00:28 A person is applying a horizontal force.

00:33 Towards right on trunk a.

00:37 Now in part a, we need to find out the maximum value of f so that no trunk moves.

00:47 Now, if we consider only trunk a, then the maximum force that can be applied on trunk a without making it move and we don't consider trunk b, then maximum force that can be can be can be applied on trunk a, f1 is equal to the force of static friction.

01:12 Because as long as the force applied force becomes more than force of static friction between the trunk and the surface, the trunk starts to move.

01:21 So the maximum limited value is equal to the force of static friction.

01:29 It should be force of friction.

01:32 And we know that force of friction is equal to new.

01:36 Into normal reaction and we know that normal reaction is equal to m1 into g say from the free body diagram we can quickly make a free body diagram this is say one of the block example it's weight m g is acting downward and normal reaction to the weight n is acting upward so n equal to mg that is why this.

02:09 Similarly, the maximum force that can be applied on trunk b without considering trunk a, f2 is equal to mu m2g.

02:23 So as both the trunks are side by side, so the maximum force that can be applied on the combination of two trunk without making the move is f equal to f1 plus f2 this is the maximum force that can be applied it means it is mu g m1 plus m2 now we have the values m1 is 0 035 g g is 9 .8 and m1 is 52 m2 is 34 so if this is 295 this is the maximum force that can be applied.

03:19 Now in part b, they are asking net force that the larger trunk a exerts on trunk b.

03:31 We need to find out the force.

03:34 Now out of the total force of 295 newton, trunk a consumes some force for itself to work against force of friction.

03:49 And that force is nothing but f1.

03:52 We have already calculated it is f1 is equal to m1g.

04:03 This is the force which is not transferred to trunk b.

04:08 It is consumed by trunk a to work against the force of friction.

04:13 It means 0 .35m .52 into 9 .8.

04:25 So it means it is 178 .36 newton.

04:33 So f1 is consumed by m1.

04:36 So remaining force is transferred from trunk a to trunk b...

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