Finding an angle given only the Coefficient of static friction


by slu1986
Tags: angle, coefficient, friction, static
slu1986
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#1
Aug30-09, 03:54 PM
P: 36
[b]1. A box sits on a horizontal wooden board. The coefficient of static friction between the box and the board is 0.22. You grab one end of the board and lift it up, keeping the other end of the board on the ground. What is the angle between the board and the horizontal direction when the box begins to slide down the board


[b]2. Relevant equations
= f / N



[b]3. I do not understand how to calculate an angle when only one variable is given. I know how to calculate the angle when you're able to calculate the sides and then use the arc tan to solve for the angle. If someone could please explain how to work out this problem I would greatly appreciate it.
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Doc Al
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Aug30-09, 04:20 PM
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Hint: What forces act on the box just before it begins to slip? What must the net force be at that point? Analyze the vertical and horizontal force components.
slu1986
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Aug30-09, 04:28 PM
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The forces that act on it are gravity 9.8 m/s^2 and static friction. The net force's must sum up to zero, but I still don't understand how to find an angle when only the coefficient of static friction is the only variable given.

Doc Al
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Aug30-09, 04:35 PM
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Finding an angle given only the Coefficient of static friction


Quote Quote by slu1986 View Post
The forces that act on it are gravity 9.8 m/s^2 and static friction. The net force's must sum up to zero, but I still don't understand how to find an angle when only the coefficient of static friction is the only variable given.
Don't forget the normal force.

Hint: Consider force components parallel to the board's surface.
slu1986
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Aug30-09, 04:39 PM
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The normal force would be the weight of the box right? But that information is not given in the problem. Only one variable is given. This is very puzzling to me b/c I don't understand how to attempt setting up a formula to solve for the angle when only 0.22 is given.
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Aug30-09, 04:49 PM
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Quote Quote by slu1986 View Post
The normal force would be the weight of the box right?
No, it depends on the angle.
But that information is not given in the problem. Only one variable is given.
Call the mass of the box "m". You won't need it.
This is very puzzling to me b/c I don't understand how to attempt setting up a formula to solve for the angle when only 0.22 is given.
Please tell me (symbolically, not with numbers):
-the normal force
-the static friction
-the component of the weight parallel to the board

Then write an equation for force components parallel to the board's surface.


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