Please help me! "A worker of a moving company places a 252-kg trunk on a piece of carpeting and slides it across the floor at constant velocity by exerting a horizontal force of 425 N on the truck."

Question

anonymous · Answer

a) What is the coefficient of kinetic friction? 
b) What happens to the coefficient of kinetic friction if another 56-kg trunk is placed on top of the 252-kg trunk? 
c) What horizontal force must the mover apply to move the combination of the two trunks at constant velocity?

anonymous · Answer

I have two formulas: 

|FN| = |Fg| = m|g| - to find the coefficient of static friction.

|FN| = |Fg| = m|g|, uK = FK/FN - coefficient of kinetic friction.

anonymous · Answer

hello.

espex · Answer

Well they tell you that it is moving at a constant velocity which means that there is no acceleration, this means that your forces equal zero.

anonymous · Answer

but the answer is 0.17...

espex · Answer

$$F_p-F_f=0 \rightarrow F_p - \mu \times F_n=0 \rightarrow \frac{F_p}{F_n}=\mu$$

anonymous · Answer

what does Fp mean?

espex · Answer

The force of the push minus the force of friction is equal to zero.

anonymous · Answer

ohh.. hmm, but I have to use the formula I stated above..

espex · Answer

Right. So you have a force pushing that is equal to 425N, you also have a force of friction which is equal to the frictional coefficient times the natural force. The natural force is mass times the force of gravity. Everything you have above.

anonymous · Answer

ohhh.. i think I get it! Thank you!

espex · Answer

You're welcome.