Finding Force by using the coefficient of friction and mass

[Alejandro Nava]

Homework Statement

A man pushes a 50 kg box up a 20-degree ramp at constant velocity. If the pushing force is parallel to the ramp, and the coefficient of friction is 0.4, how much force is he pushing with?

Homework Equations

u=Fs, max/Fn Not sure if this equation is needed.

The Attempt at a Solution

I tried to solve the problem but I ended up getting the frictional force instead of the force he is pushing with.

 

[gneill]

Show us the details of what what you've tried so far.

 

[CWatters]

Make a free body diagram.

 

[neilparker62]

Forces parallel to ramp mgsin(θ) and μFn. The man is pushing against weight component down the slope and friction.

 

[kuruman]

Forces parallel to ramp mgsin(θ) and μFn.

Is that all?

 

[ZapperZ]

The Attempt at a Solution

I tried to solve the problem but I ended up getting the frictional force instead of the force he is pushing with.

What you wrote here does not qualify as "attempt at a solution". You told us you couldn't do it (which is automatic when someone posts something in this subforum), but without seeing what you actually did, there is no way anyone can diagnose what you did wrong.

We are not allowed to give you outright solutions. We are, however, allowed to help you find the solution. And to do that, we need to know what you actually did. So show your work, especially your free-body diagram, as has been suggested.

Zz.

 

[neilparker62]

Is that all?

Yes.

 

[kuruman]

Yes.

What about the man pushing?

 

[neilparker62]

As mentioned in my post, the man is pushing against the above-mentioned forces and there will be zero resultant force since the block does not accelerate.

 

[kuruman]

I agree that the resultant force is zero. I disagree about the number of forces acting on the block. I can count 3 in the direction parallel to the incline that should add to zero: pushing force by man, friction by incline, component of gravity. How many are there according to your count?

 

[neilparker62]

Quite right - there are 3 forces in equilibrium so the OP will need to add two of them to obtain the third. Or alternatively add all 3 and set equal to zero with the pushing force being the unknown.

 

[kuruman]

Quite right - there are 3 forces in equilibrium so the OP will need to add two of them to obtain the third. Or alternatively add all 3 and set equal to zero with the pushing force being the unknown.

You got it!

 

[neilparker62]

You got it!

More importantly I hope the OP did too!

 

[Alejandro Nava]

Thank you all for trying but I ended up going to the teacher to ask how to do the problem and I understand how to do the question. What I did wrong was that when I was trying to find the force, I used COH instead of TOA. Sorry for the inconvenience.

 

[neilparker62]

Show us the details of what what you've tried so far.

Perhaps this was the correct response to the OP's query! I'm not sure what either COH nor TOA mean.

 

[jbriggs444]

Perhaps this was the correct response to the OP's query! I'm not sure what either COH nor TOA mean.

"COH" and "TOA" are apparently intended to be part of "SOHCAHTOA", which is a mnemonic for remembering how sine, cosine and tangent are found in a right triangle.

http://mathworld.wolfram.com/[URL="...mingly-simple-conceivably-complex/"]SOHCAHTOA.html[/URL]

 

[neilparker62]

"COH" and "TOA" are apparently intended to be part of "SOHCAHTOA", which is a mnemonic for remembering how sine, cosine and tangent are found in a right triangle.

http://mathworld.wolfram.com/[URL="...mingly-simple-conceivably-complex/"]SOHCAHTOA.html[/URL]

Thanks for the clarification.