Dochopeful13

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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
How did they get this answer?
 

Danny L

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Did the answer actually include the sqroot of F? This is what I got:

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

I think it's weird that they asked you to take that extra step and solve the trig/simplify the fraction
 
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Dochopeful13

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Thank you for your help. Math is not my strong point. Can you please tell me how you got from

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

Danny L

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Thank you for your help. Math is not my strong point. Can you please tell me how you got from

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

Yeah, so if you have a fraction like (x+y)/z, you can split the fraction into (x/z)+(y/z). It's basically the reverse of adding two fractions with the same denominator (so for example with actual numbers, 2/6 + 2/6 = (2+2)/6 or 4/6)
 
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Dochopeful13

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Yeah, so if you have a fraction like (x+y)/z, you can split the fraction into (x/z)+(y/z). It's basically the reverse of adding two fractions with the same denominator (so for example with actual numbers, 2/6 + 2/6 = (2+2)/6 or 4/6)
Thank you so much.Would you mind explaing where the sq root comes into play?
 

Danny L

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Thank you so much.Would you mind explaing where the sq root comes into play?

cos45 and sin45 is equal to the sqroot(2)/2 (unit circle). Based on the answer you gave, it looked like they solved the cos and sin (although I don't think the actual MCAT would ask you to do that)

Was 2F^(1/2)/mg supposed to to be 2F/(mg*sqroot(2))?
 
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Dochopeful13

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cos45 and sin45 is equal to the sqroot(2)/2 (unit circle). Based on the answer you gave, it looked like they solved the cos and sin (although I don't think the actual MCAT would ask you to do that)

Was 2F^(1/2)/mg supposed to to be 2F/(mg*sqroot(2))?
I really appreciate all of your help. The answer was sqrt 2f/mg
 

Dochopeful13

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cos45 and sin45 is equal to the sqroot(2)/2 (unit circle). Based on the answer you gave, it looked like they solved the cos and sin (although I don't think the actual MCAT would ask you to do that)

Was 2F^(1/2)/mg supposed to to be 2F/(mg*sqroot(2))?



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
 

Danny L

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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

Fnet is equal to 0, which is why you can set F (the variable given to applied force by the problem) equal to the resisting forces of gravity and friction

F-F(gravity)-F(friction)=Fnet=0
F-mgsin45-umgcos45=Fnet=0
 
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