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Static Friction and minimum force

by anne921
Tags: force, friction, minimum, static
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anne921
#1
Jan9-13, 12:30 PM
P: 10
1. The problem statement, all variables and given/known data
The coefficient of static friction between the box (6.75 kg) and the ramp (30 degree incline) is 0.500. What is the magnitude of the minimum force, F, that must be applied to the box perpendicularly to the ramp to prevent the box from sliding?


2. Relevant equations



3. The attempt at a solution
The key says the answer to this question is 124N but I am unable to reach that answer. I think I am missing something and/or using the wrong equation. Please help me!
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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gneill
#2
Jan9-13, 01:09 PM
Mentor
P: 11,678
Quote Quote by anne921 View Post
1. The problem statement, all variables and given/known data
The coefficient of static friction between the box (6.75 kg) and the ramp (30 degree incline) is 0.500. What is the magnitude of the minimum force, F, that must be applied to the box perpendicularly to the ramp to prevent the box from sliding?


2. Relevant equations



3. The attempt at a solution
The key says the answer to this question is 124N but I am unable to reach that answer. I think I am missing something and/or using the wrong equation. Please help me!
Hello anne921, Welcome to Physics Forums.

Can you show the work you did on your attempt so that we can see how to help?
anne921
#3
Jan9-13, 01:49 PM
P: 10
Yes - the simplest approach would be to use the formula that equates the coefficient of static friction to static friction force divided by the normal force.

Static Friction Force = 0.500 X mgcos30 = 28.7 N

Another thing I thought made sense is to add the force due to the mass of the box (mgcos30) to the X component of the force of gravity (mgsin30):

Maximum Force - mgcos30 + mgsin30 = 57.29N + 33.11N = 90.4N

Both of these are answer choices (multiple choice) for the question.

gneill
#4
Jan9-13, 02:06 PM
Mentor
P: 11,678
Static Friction and minimum force

Quote Quote by anne921 View Post
Yes - the simplest approach would be to use the formula that equates the coefficient of static friction to static friction force divided by the normal force.

Static Friction Force = 0.500 X mgcos30 = 28.7 N
Okay, that would be the static friction force due to the normal component of the weight of the box, and not including any additional normal force due to the force F that is to be applied.
Another thing I thought made sense is to add the force due to the mass of the box (mgcos30) to the X component of the force of gravity (mgsin30):

Maximum Force - mgcos30 + mgsin30 = 57.29N + 33.11N = 90.4N
I don't see that making sense, since those forces are perpendicular to each other; One is directed into the plane of the ramp, the other is parallel to the ramp.

What you should do is compare the "downslope" force due to the weight of the box to the static friction force. If the friction force is less than the downlope force, the box will slide.

Now, adding force F to press the box more firmly against the slope is going to increase the normal force, and hence increase the friction. But it won't increase the downlsope force which is due to the weight of the box alone. Can you write an expression for the "new" friction force when F is applied?
anne921
#5
Jan9-13, 02:29 PM
P: 10
OK - so am I misinterpreting the question?? Is there an additional force that must be applied to keep the box from slipping? I am afraid that I don't know what you are saying. I thought that adding the downward slope force in was a mistake but am really confused as to what to do...
gneill
#6
Jan9-13, 02:39 PM
Mentor
P: 11,678
Quote Quote by anne921 View Post
OK - so am I misinterpreting the question?? Is there an additional force that must be applied to keep the box from slipping? I am afraid that I don't know what you are saying. I thought that adding the downward slope force in was a mistake but am really confused as to what to do...
The problem statement read, in part, "What is the magnitude of the minimum force, F, that must be applied to the box perpendicularly to the ramp to prevent the box from sliding?"

So some external force F is to be applied, pushing the box more firmly against the slope. The object is to find the magnitude of this force F.



In the figure, the forces labelled in black font are the ones due to gravity acting on the box. The "new" force F is in blue. The force due to friction ##f_f## is also indicated in blue.
Attached Thumbnails
Fig1.gif  
anne921
#7
Jan10-13, 08:56 PM
P: 10
OK - I am very new to this and am trying to explain it to others. I seem to be missing a key point here in the concept. I understand what you are saying but have no idea how to calculate this additional force. Please get me started.
gneill
#8
Jan10-13, 09:34 PM
Mentor
P: 11,678
Start by considering the box alone, without the additional force F involved. Can you determine the downslope force fdg, the normal force fng and the resulting frictional force resulting from fng?
anne921
#9
Jan10-13, 10:56 PM
P: 10
Yes, OK - Fdg would be mgsinθ ; Fng would be mgcosθ ; and the frictional force due to Fng would be equivalent to the normal force since the box isn't moving...I am not positive about this last one but I am thinking since there is no motion the frictional force is that due to the box weight only.
gneill
#10
Jan10-13, 11:03 PM
Mentor
P: 11,678
Quote Quote by anne921 View Post
Yes, OK - Fdg would be mgsinθ ; Fng would be mgcosθ ; and the frictional force due to Fng would be equivalent to the normal force since the box isn't moving...I am not positive about this last one but I am thinking since there is no motion the frictional force is that due to the box weight only.
Yes to the first two. The frictional force is a portion of the normal force, that portion being determined by the coefficient of friction.

At this point, without the additional force F being applied, we can't say that the box won't move; We have to compare the frictional force to the downslope force. If the downslope force exceeds the frictional force it will start to move. So the question then becomes, how much additional normal force (F) is required to boost the frictional force to the point that it just balances the downslope force.
anne921
#11
Jan23-13, 09:57 PM
P: 10
I am still struggling with this one. I was able to solve it I think...

the parallel component of gravity is mgsin30 which is 33.1 N; so the frictional force must balance this out so Fn = Ff/0.5 = 66.15N. Y component of gravity is mgcos30 which is 57.4N. The additional component of force necessary would then be the difference between the normal force and the y component of gravity - 8.8N. My key doesn't give this as an answer choice and I don't have the expertise in physics to know if I am correct. Is my solution correct?
gneill
#12
Jan23-13, 10:15 PM
Mentor
P: 11,678
Your result looks okay to me.
anne921
#13
Jan25-13, 01:34 PM
P: 10
Thanks!


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