What is the critical value of mu_s in this static friction problem?

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The discussion centers on calculating the critical value of the static friction coefficient (mu_s) for a box being pushed up an inclined ramp. The equations of motion are established by resolving forces parallel and perpendicular to the incline, leading to two key equations that involve the applied force (F), gravitational force, and normal force (N). The maximum static friction is defined as f_s = mu_s * N, where N is derived from the weight of the box. By rearranging the equations, the critical value of mu_s can be expressed as mu_s = (m*g - F*Gcos(alpha)) / (F*Gsin(alpha)). This value indicates the threshold above which the box will not move up the ramp.
Lalo1985
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Hi, I'm having trouble solving this problem:

A woman attempts to push a box of books that has mass up a ramp inclined at an angle (alpha) above the horizontal. The coefficients of friction between the ramp and the box are (mu_k) and (mu_s). The force F applied by the woman is horizontal.

If (mu_s) is greater than some critical value, the woman cannot start the box moving up the ramp no matter how hard she pushes. Calculate this critical value of (mu_s).

I know that the maximum value of mu_s is f_mu/N. So, using F(y) = ma(y), I got: N - F*Gcos(alpha) - F*Gsin(alpha) = 0. Therefore, making N = F*Gcos(alpha) + F*Gsin(alpha). That's it. I don't know what to do next. Any ideas?
 
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You have to resolve F in two directions: parallel to the incline and perpendicular to it. This will give you two equations, both involving F and one involving the normal reaction N, friction force f and acceleration a. Since the mass (of books) does not move up or down the incline, the acceleration a = 0. So this problem reduces to

\sum F_{x} = 0

\sum F_{y} = 0

where x and y are the directions parallel and perpendicular to the incline.

You should get the following equations as a result

F\cos\alpha - mg\sin\alpha - f_{s} = 0
N - mg\cos\alpha - F\sin\alpha = 0

Solve them to get the value of \mu_{s} using the fact that f_{s} = \mu_{s}N.

Hope that helps...

Cheers
Vivek

EDIT--The above equations are valid iff F is applied in the horizontal direction (i.e. in a direction parallel to the base of the incline).
 
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To solve this problem, we can use the concept of static friction and its relationship with the coefficient of friction. When an object is at rest, the force of static friction is equal to the applied force, up to a maximum value determined by the coefficient of friction. In this case, the applied force is the horizontal force F applied by the woman, and the maximum value of static friction is mu_s*N, where N is the normal force exerted by the ramp on the box.

To find the critical value of mu_s, we need to determine the maximum value of N. From the free body diagram, we can see that the normal force N is equal to the weight of the box, which is given by m*g, where m is the mass of the box and g is the acceleration due to gravity.

Therefore, the maximum value of N is m*g. Substituting this into our equation for N, we get:

m*g = F*Gcos(alpha) + F*Gsin(alpha)

Now, we can solve for the critical value of mu_s by rearranging the equation:

mu_s = (m*g - F*Gcos(alpha)) / (F*Gsin(alpha))

This is the critical value of mu_s that the woman cannot exceed in order to start the box moving up the ramp. Any value of mu_s greater than this will result in the box remaining at rest.

I hope this helps you solve the problem. Remember to always carefully consider the forces acting on an object and their relationships in order to solve physics problems. Good luck!
 
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