TPRH Chapter 3 FSQ #4

[aspiringdoc09]

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Question: A box of mass m is sitting on an incline of 45 degress 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?

a) [sqrt2*F / mg] - 1
b) [sqrt2*F / mg]
c) [sqrt2*F / mg] + 1
d) [2F / mg] - 1

How did they arrive at this answer. I must be doing something wrong both mathematically and solving the physics. I made a stupid mistake and forgot that frictional force + gravitational force will act downwards. I was so use to frictional force acting downward and the pushing force acting up the ramp. Therefore, it led me to take the difference between the two. After looking at explanation, now I can't figure out how they are manipulating the variables to get choice A. Where did the squareroot of 2 come from? Help please.

 

[dmf2682]

Question: A box of mass m is sitting on an incline of 45 degress 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?

a) [sqrt2*F / mg] - 1
b) [sqrt2*F / mg]
c) [sqrt2*F / mg] + 1
d) [2F / mg] - 1

How did they arrive at this answer. I must be doing something wrong both mathematically and solving the physics. I made a stupid mistake and forgot that frictional force + gravitational force will act downwards. I was so use to frictional force acting downward and the pushing force acting up the ramp. Therefore, it led me to take the difference between the two. After looking at explanation, now I can't figure out how they are manipulating the variables to get choice A. Where did the squareroot of 2 come from? Help please.

I get choice b and would be curious how a is the answer. Thought I knew physics but it's been 12 years so I'm a tad rusty 🙂

 

[dmf2682]

I get choice b and would be curious how a is the answer. Thought I knew physics but it's been 12 years so I'm a tad rusty 🙂

Oh lol, Nevermind I got it.

Draw a picture and free body diagram.

You have gravity going down. Use trig to get components parallel
And normal to the incline. So normal force is mg/sqrt2, and parallel force is the same. Force from friction is mu N, or mg*mu/sqrt2. This equals your net force parallel to the incline which is force f minus the parallel component from gravity. Plug and chug gets you your answer.

 

[aspiringdoc09]

Oh lol, Nevermind I got it.

Draw a picture and free body diagram.

You have gravity going down. Use trig to get components parallel
And normal to the incline. So normal force is mg/sqrt2, and parallel force is the same. Force from friction is mu N, or mg*mu/sqrt2. This equals your net force parallel to the incline which is force f minus the parallel component from gravity. Plug and chug gets you your answer.

I don't see how you are getting normal force = mg/sqrt2. I know because this is a special triangle (45-45-90) that two legs will be the same; therefore, normal force = parallel force. I don't understand why you are dividing mg/sqrt2 instead of multiplying mg*sqrt2?

 

[Pisiform]

F = mgSin 45 + (mu) mg Cos 45

F = mg/sqrt2 + (mu)mg/sqrt2

mu = (F - mg/sqrt2) * sqrt2/mg

mu = Fsqrt2/mg - 1

 

[aspiringdoc09]

Oh lol, Nevermind I got it.

Draw a picture and free body diagram.

You have gravity going down. Use trig to get components parallel
And normal to the incline. So normal force is mg/sqrt2, and parallel force is the same. Force from friction is mu N, or mg*mu/sqrt2. This equals your net force parallel to the incline which is force f minus the parallel component from gravity. Plug and chug gets you your answer.

F = mgSin 45 + (mu) mg Cos 45

F = mg/sqrt2 + (mu)mg/sqrt2

mu = (F - mg/sqrt2) * sqrt2/mg

mu = Fsqrt2/mg - 1

I'm sorry. I feel so dumb! Stupid mistakes as always. I had fixated in my mind that the sides were each sqrt 2 and the hypotenuse 1, but it should be the other way around. And if I use trig then I would see that both the sin 45 and cos 45 are 1/sqrt2. Also, because both frictional and gravitational forces are in the same direction, we add them. It took me a little bit to see what I was doing wrong now that my brain is actually working. I need to stop making mistakes like this. It will kill my MCAT score. Thanks for your help.

 

[4 lyfe]

I think there is something wrong with this question... first of all sin 45 = cos 45 = (2^0.5)/2 not just 2^0.5.... so here is my work and I'm pretty sure its right, unless you guys (and gals) can show me otherwise... here we go!

F= force, FS= static force

F- FS - mu mg cos 45 - mg sin 45 = 0

mu =( F - mg sin 45 ) / mg cos 45

mu = F / mg cos 45 - mg sin 45/ mg cos 45

mu = { F / mg (2^0.5/2) } - 1

mu = (2F / mg 2^0.5 ) - 1

I believe this is the correct answer... again punch sin 45 and cos 45 into your calculator and you'll see that they equal 0.707 something, which is equal to 2 ^ 0.5 / 2...

 

[dmf2682]

I think there is something wrong with this question... first of all sin 45 = cos 45 = (2^0.5)/2 not just 2^0.5.... so here is my work and I'm pretty sure its right, unless you guys (and gals) can show me otherwise... here we go!

F= force, FS= static force

F- FS - mu mg cos 45 - mg sin 45 = 0

mu =( F - mg sin 45 ) / mg cos 45

mu = F / mg cos 45 - mg sin 45/ mg cos 45

mu = { F / mg (2^0.5/2) } - 1

mu = (2F / mg 2^0.5 ) - 1

I believe this is the correct answer... again punch sin 45 and cos 45 into your calculator and you'll see that they equal 0.707 something, which is equal to 2 ^ 0.5 / 2...

Your fancy calculator should also tell you that 2/sqrt2 = sqrt2. We've the same answers.

Multiply sqrt2/2 by one, which can be expressed as sqrt2/sqrt2. You get (sqrt2)^2 in the numerator and 2sqrt2 in the denominator. Numerator equals 2, cancels the 2 in the denominator, so the fraction becomes 1/sqrt2.

 

[4 lyfe]

you are right mate thanks !!! my stupidity... forgive what have may appear to be arrogance...

 

[dmf2682]

you are right mate thanks !!! my stupidity... forgive what have may appear to be arrogance...

No problem. I was taught in high school that it was poor form to leave a radical in the denominator, so when I first did the problem I had it as you show it, but it's good to be able to switch around, poor form or not 🙂