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Forces  finding minimum mass given coefficients of friction 
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#1
Jul1010, 12:05 PM

P: 78

1. The problem statement, all variables and given/known data
A rope exerts a force of magnitude of 21 N, at an angle 31 degrees above the horizontal, on a box at rest on a horizontal floor. The coefficients of friction between the box and the floor are s=0.55 and k=0.50 . The box remains at rest. Determine the smallest possible mass of the box. 2. Relevant equations Fs= s*Fnormal , Fk=k*Fnormal F=ma 3. The attempt at a solution I've attached my system diagram to the post. Using vector components, I find that Fappx=18 N and Fappy=10.8N. That's all I know how to do here :/. I have to solve for mass, but I'm not sure how the physics is working. Can someone start me off on how to solve this question? 


#2
Jul1010, 12:17 PM

P: 621




#3
Jul1010, 12:29 PM

P: 621

You have calculated Fx, or the applied horizontal force, so thats done. How are you going to calculate Fs? In other words, can you write an equation to calculate Fs? I bet the equation you write will have one unknown, m...



#4
Jul1010, 01:09 PM

P: 78

Forces  finding minimum mass given coefficients of friction
Fs = s*Fn Fs = 0.55 * (9.8m+10.8) ...I think this is it? I'm lost with how to manipulate Fk though. 


#5
Jul1010, 01:16 PM

P: 621




#6
Jul1010, 01:20 PM

P: 621

What does the normal for Fn equal in this situation? I think you may have a sign problem.



#7
Jul1010, 01:22 PM

P: 78

If the object is at rest then Fs= Fapp, like you said before.
and Fapp = Fappx (horizontal component of the applied force). So then 18 = 0.55 * (9.8m+10.8) m = 2.03 kg 


#8
Jul1010, 01:23 PM

P: 78

Bah, the answer is 4.4kg.
Did I do something wrong? 


#9
Jul1010, 01:25 PM

P: 78

This is what I have as the normal force:
Fn = 9.8m+10.8 It should be a positive value, no? 


#10
Jul1010, 01:30 PM

P: 621




#11
Jul1010, 02:00 PM

P: 78

So ...the sign is flipped! I guess that makes sense because the F vertical is pointing downwards. But then, I didn't make 9.8m/s^2 negative, even though its the downward acceleration of gravity. I get confused about sign flipping all the time, but okay, thank you! 


#12
Jul1010, 02:05 PM

P: 621

Well the push back from the table on the block has to be less than mg if you are also pulling up on the block with a string. The Normal must be less. The normal force would be more than mg if you pushed down on a block with your finger. The table would push back with mg + force of finger. I hope this makes sense. 


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