How Much Force Does the Larger Trunk Exert on the Smaller Trunk?

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Sace Ver
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Homework Statement



Two trunks sit side by side on the floor. The larger trunk (52kg) 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 floor is 0.35. Calculate the force the larger trunk exerts on the smaller trunk?

mass of larger trunk = 52kg
mass of smaller trunk = 34kg
uk = 0.35

Homework Equations



uf = Ff/FN
Fnet = sum of all forces

The Attempt at a Solution



FSmax = usFN
FSmax = (0.35)(86kg)(9.8N/kg)
FSmax = 290N

FN = mg
FN = (52kg)(9.8N/kg)
FN = 509.6N

Not quite sure if doing right steps or what to do next?[/B]
 
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Mister T said:
Are you sure you've given us the problem? It seems something is missing.
The magnitude of the maximum force the person can exert without moving either trunk is 290N! And that's basically it!
 
Sace Ver said:
The magnitude of the maximum force the person can exert without moving either trunk is 290N!

Agreed, but nowhere in the statement of the problem is it mentioned that the maximum force is being applied. Neither do they state that the trunks aren't moving.

Anyway, if you make those assumptions the next step is to look at the horizontal force applied to the 34 kg trunk. Or to the horizontal forces applied to the 52 kg trunk. Drawing the free body diagrams is a big help.
 
Mister T said:
Agreed, but nowhere in the statement of the problem is it mentioned that the maximum force is being applied. Neither do they state that the trunks aren't moving.

Anyway, if you make those assumptions the next step is to look at the horizontal force applied to the 34 kg trunk. Or to the horizontal forces applied to the 52 kg trunk. Drawing the free body diagrams is a big help.
I'm pretty sure the trunks are not moving meaning acceleration would be zero and so would Fnet? And I drew the FBD and just wondering would FT and FF cancel each other out?
 
Mister T said:
What are FT and FF?
I think I made an error on my FBD, because I have yet to find FT but would FF be 290N?
And I'm looking for FT right? And I'm just wondering if FT equals FSmax?
 
Mister T said:
What I meant was, how are you defining FT and FF?
I think I figured it out

Fsmax = usF N
Fsmax = (0.35)(34kg)(9.8N/kg)
Fsmax = 120N
 
Sace Ver said:
I think I figured it out

Fsmax = usF N
Fsmax = (0.35)(34kg)(9.8N/kg)
Fsmax = 120N
Right, though as E=mc2 noted you have to fill in two missing pieces of information to arrive at that: that the trunks are not moving, but if the force between the trunks were any greater they would be. In the question as given, the pusher could be just breathing heavily against the larger trunk.