Question involving Newton's Laws and Friction

AI Thread Summary
The discussion revolves around a scenario where a worker pulls a 10 kg crate on a rough floor with a force of 45N. The coefficients of static and kinetic friction are 0.5 and 0.3, respectively. The normal force acting on the crate is calculated to be 98N, leading to a maximum static friction force of 49N. Since the pulling force of 45N is less than the maximum static friction, the frictional force matches the pulling force. Consequently, the frictional force exerted by the surface is 45N.
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Homework Statement



A worker pulls horizontally on a rope that is attached to a 10 kg crate that is resting on a rough floor. The coefficients of static and kinetic friction are 0.5 and 0.3 respectively. The worker pulls with a force of 45N. What is the frictional force exerted by the surface?

We did this in class and the answer was 45N but I do not understand why that is. Can someone explain?
 
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A worker pulls horizontally on a rope that is attached to a 10 kg crate that is resting on a rough floor. The coefficients of static and kinetic friction are 0.5 and 0.3 respectively. The worker pulls with a force of 45N. What is the frictional force exerted by the surface?

We did this in class and the answer was 45N but I do not understand why that is. Can someone explain?

So vertical direction of the pull does not need to be taken into account, therefore the normal force = weight.

The weight = mg or (10kg)(9.8) = 98 N

so the normal force = 98N

since f = (coefficient) * N

f = .5 * 98 = 49N, so the worker would need to pull with a force GREATER than 49 N in order to start accelerating the box, but since he pulls with only 45N, the force of friction will be equal to the pulling force..

so therefore f = force of pull = 45 N

so f = 45N
 
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