Stuck on TPR Physics Problem

[Dochopeful13]

I have question from TPR Physics.

1. A box of mass m is sitting on an incline of 45 degrees and it requires an applied force F up the incline to get the box to begin to move. What is the maximum coefficient of static friction?

I know that the parallel component of the gravitational force is mgsin45 and the perpendicular component is mgcos45. The maximum static frictional force would be (normal force mgcos45)*the static coefficient. The question asked about when the box is just about to move so the upward and downward forces are equal and static friction is at a max. But the explanation states F=static coefficient*mgcos45+static coefficient*sin45 and from there you can solve for the coefficient. Why is the static coefficient multiplied to sin45 is it bc both gravitational and static forces are acting in the same direction opposing motion?

answer is {2F^(1/2)/(mg)}-1

Is this a typo?

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[NarutoMD]

I think this "F=static coefficient*mgcos45+static coefficient*sin45" may be a typo. I don't see how they could manipulate the set up to get
static coefficient*sin45.
What I got is: F= mgsin45+mgcos45*static coefficient ( This will be my set up since both the parallel component and friction force oppose the applied force)
static coefficient= (F - mgsin45)/mgcos45-> F/(mgsin45)-1 (since sin45=cos 45)
so static coefficient = F/ (mg*sqrt(2)/2) - 1
We can manipulate the sqrt2/2 by multiplying by 2/sqrt(2) to F/(mg*sqrt(2)/2) to get (2/sqrt2)F/(mg) - 1
So static coefficient = (2/sqrt 2)F/(mg) - 1 ( I dont know how they could get sqrt F); I am not really good at manipulating things so this is as far as I could go. Maybe someone else has a better idea.

 

[5words]

Might or might not be an error, Because the Force of static friction is able to supply/ provide any force. Thus, if the box isn't moving, it's fair to imply that the Block is providing mgsing45 *us Force of static friction. However, since they asked for the maximal, the formula to use in this case, is mgcos45*us . I understand how forces are vector, but not getting why they are adding them. It's either F of static friction = F applied when box is not moving or F = N*us when Frictional force is maximal.

So would you mind posting the problem?

 

[Dochopeful13]

Here is the problem worked out

Might or might not be an error, Because the Force of static friction is able to supply/ provide any force. Thus, if the box isn't moving,

Might or might not be an error, Because the Force of static friction is able to supply/ provide any force. Thus, if the box isn't moving, it's fair to imply that the Block is providing mgsing45 *us Force of static friction. However, since they asked for the maximal, the formula to use in this case, is mgcos45*us . I understand how forces are vector, but not getting why they are adding them. It's either F of static friction = F applied when box is not moving or F = N*us when Frictional force is maximal.

So would you mind posting the problem?

This what I am told is right.

u=static coefficient, F=force given in the problem

F=mgsin45+umgcos45
u=(F-mgsin45)/mgcos45
u=(F/mgcos45)-(mgsin45/mgcos45)

sin45=cos45=sqrt(2)/2

u=[2F/(mg*sqrt(2))]-1

 

[5words]

oh, nvm then .. BTW just read your other post, is F under the square root as well?

 

[Dochopeful13]

oh, nvm then .. BTW just read your other post, is F under the square root as well?

Yes it is. What confuses me isn't Fnet equal to zero in this problem since this is in equilibrium? If so why can't we set everything to zero? Since friction is not a conservative force wouldn't we subtract friction from mgsin? Could we set up the problem this way

0=Mgsin45-umgcos45

 

[5words]

Yes it is. What confuses me isn't Fnet equal to zero in this problem since this is in equilibrium? If so why can't we set everything to zero? Since friction is not a conservative force wouldn't we subtract friction from mgsin? Could we set up the problem this way

0=Mgsin45-umgcos45

hence why i asked, if it was possible to post the problem itself (nvm, you posted it). But to answer your question, it depends, i saw you mentioned applied force ealier if that's the case then i'll go over it in 2. But

Fnet = ma , still Fnet is the sum of forces acting on the object so i guess you cant set it up as Fnet = mgsingx -/+ uN , and if the object isn't moving then that Fnet =0 or Fnet = ma if the object is moving.

If there is a there force being applied to the system, then your Fnet becomes Faplied + mgsinX +/- uN. And because, you most of the time have static friction when the object is stationary thus it follows that Fapplied + mgsinX +/- uN = 0

So, it should be -mgsinX - uN + F(applied to get the box up) =Fnet because you;are asking the coefficient of static friction, it's safe to assume that the object is stationary thus Fnet = 0 .. Guarantee that it's a typo

But the explanation states F=static coefficient*mgcos45+static coefficient*sin45 and from there you can solve for the coefficient. Why is the static coefficient multiplied to sin45 is it bc both gravitational and static forces are acting in the same direction opposing motion?

I guess here they swapped mg with us.

 

[NarutoMD]

Yes it is. What confuses me isn't Fnet equal to zero in this problem since this is in equilibrium? If so why can't we set everything to zero? Since friction is not a conservative force wouldn't we subtract friction from mgsin? Could we set up the problem this way

0=Mgsin45-umgcos45

This will work only if there is no other forces acting on it. Here, there is a force acting up on an object on an incline. Friction will always act against an applied force, meaning in the opposite direction. This means that friction will work in the same direction as gravity (since gravity is pulling down on the object down the incline).

 

[Dochopeful13]

This will work only if there is no other forces acting on it. Here, there is a force acting up on an object on an incline. Friction will always act against an applied force, meaning in the opposite direction. This means that friction will work in the same direction as gravity (since gravity is pulling down on the object down the incline).

Ahh makes total sense thank you. Is it fair to say if the applied force was applied downwards with gravity then the problem would be f=mgcos-mgsin or does gravity and friction always work against the applied force together?

 

[5words]

Ahh makes total sense thank you. Is it fair to say if the applied force was applied downwards with gravity then the problem would be f=mgcos-mgsin or does gravity and friction always work against the applied force together?

Yeah, you can say that.

 

[NarutoMD]

Ahh makes total sense thank you. Is it fair to say if the applied force was applied downwards with gravity then the problem would be f=mgcos-mgsin or does gravity and friction always work against the applied force together?

No, gravity does not always work against the applied force. Only when the applied force is in the opposite direction (upward) to gravity will gravity opposes the applied force. This will make more sense to you if you draw out a diagram. Gravity always points downward, so any force that points in the opposite direction will be working against gravity. What you said about the applied force being pointed downward with gravity is correct.