Is my solution good? - kinetic friction

AI Thread Summary
The discussion revolves around the effects of kinetic friction on a block being pulled with a force F in two scenarios: one with a horizontal force and the other at a 60-degree angle. It is clarified that kinetic friction depends on the normal force, which varies between the two situations due to the angle of the applied force. In the first scenario, the normal force equals the weight of the block, while in the second, it is reduced by the vertical component of the applied force. The calculations presented suggest that the coefficient of kinetic friction (u) is crucial for determining the motion in both cases. Overall, the initial assumption that motion could be uniform in both situations is incorrect due to the differing normal forces affecting friction.
irrehaare
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in first situation block is pulled with force F .

in second situation we have the same block with the same force F but under angel 60 degrees.

it is possible that in both situation motion is uniform. ?? (we have friction)

so imo it isn't possible because The block is in the both situation in motion, so we have kinetic friction and it's depends on a gravitacion force, so in both situation should be the same.

My thought is correct or no? somebody could give me an answer ?
 
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Careful: The kinetic friction depends on the normal force, not just the weight of the block. When the force F is applied at an angle, the normal force changes.
 
Thx now i correctmy solution
i have somthing like this:

F-uQ=0
and
F/2 - u(Q-Fcos60)=0

my result is:
u=3/cos60
so the solution depends on the u (coefficient of kinetic friction)

Now am J correct ?
 
Not sure what you are doing here, or what you are trying to solve for.

In situation 1, where the applied force is horizontal:
- the normal force equals the weight of the object N = mg
- the net horizontal force is F - \mu N

In situation 2, where the applied force is at an angle with respect to the horizontal:
- the normal force equals N = mg - F\sin \theta
- the net horizontal force is F \cos \theta - \mu N

(I'm moving this to the homework help section.)
 
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