Coefficient of static friction of crate

AI Thread Summary
To determine the coefficient of static friction for a 200N crate on a 20-degree ramp, it is essential to analyze the forces acting on the crate, which include gravitational force, normal force, and frictional force. The normal force can be calculated using the weight of the crate and the angle of the ramp. The equation T = N/cos(theta) is relevant, but it needs to be applied correctly to find the normal force. The static friction coefficient is derived from the ratio of the frictional force to the normal force, leading to the answer of 0.47. Understanding the components of these forces in an efficient coordinate system is crucial for solving the problem accurately.
collegegirl
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Homework Statement


A 200N crate rests on a ramp, the angle is 20 degrees with the horizontal. What is the coefficient of static friction between the crate and ramp surfaces


Homework Equations


T=N/cos theta


The Attempt at a Solution


200/cos25 = 219

I don't know where to go from here. The answer is .47 Please show me a direction to take.!

 
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collegegirl said:

Homework Statement


A 200N crate rests on a ramp, the angle is 20 degrees with the horizontal. What is the coefficient of static friction between the crate and ramp surfaces


Homework Equations


T=N/cos theta


The Attempt at a Solution


200/cos25 = 219

I don't know where to go from here. The answer is .47 Please show me a direction to take.!

What forces are acting upon the object? (Hint: There are three forces and one of these forces has components if you choose an efficient coordinate system.)
 
Would the three forces be gravity, the weight of the crate on the ramp, and the ramp pushing up on the crate? If so, the only one I have numbers for is gravity??
 
collegegirl said:
Would the three forces be gravity, the weight of the crate on the ramp, and the ramp pushing up on the crate? If so, the only one I have numbers for is gravity??

Weight and gravity are related by \vec{w}=m\vec{g}. The other two forces are going to be the frictional force (a function of the normal force) and the normal force, itself, a function of weight. So, where are your x- and y-axes going to lie?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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