Is my solution good? - kinetic friction

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Homework Help Overview

The discussion revolves around a problem involving kinetic friction and the effects of an applied force at different angles on a block's motion. Participants explore the implications of friction in two scenarios: one where the force is applied horizontally and another where it is applied at a 60-degree angle.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to determine if uniform motion is possible in both scenarios despite the presence of kinetic friction. They question whether their reasoning about the dependence of kinetic friction on gravitational force is correct. Other participants raise points about the role of the normal force and its changes due to the angle of the applied force.

Discussion Status

Some participants have provided clarifications regarding the relationship between the applied force, normal force, and kinetic friction. There is ongoing exploration of the calculations related to the coefficient of kinetic friction, but no consensus has been reached on the correctness of the original poster's solution.

Contextual Notes

Participants are navigating the complexities of how the angle of the applied force affects the normal force and, consequently, the kinetic friction. The discussion is framed within the constraints of homework expectations, focusing on understanding rather than providing direct solutions.

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 [itex]N = mg[/itex]
- the net horizontal force is [itex]F - \mu N[/itex]

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

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

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