Hi I cannot seem to figure out the answer to part a and part b A large crate wit
ID: 1609290 • Letter: H
Question
Hi I cannot seem to figure out the answer to part a and part b
A large crate with mass m rests on a horizontal floor. The coefficients of friction between the crate and the floor are mu_s and mu_k. A woman pushes downward at an angle theta below the horizontal on the crate with a force F vector. What magnitude of force F vector is required to keep the crate moving at constant velocity? F = If mu_s is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. Calculate this critical value of mu_s.Explanation / Answer
F cos is the horizontal force pushing the object horizontally.
F sin is the vertical force pushing the object into the floor,
The weight of the object also pushes the object into the floor.
The floor prevents the box from going into the floor.
There is no vertical motion for the box.
If somehow, the box is set into motion with constant speed, then
The frictional force that acts horizontally is k (Mg + F sin )
Since the box has constant speed, there is no acceleration and hence the applied horizontal force equals the opposite frictional force.
F cos = k (Mg + F sin )
F cos - k F sin = k Mg
F (cos - k sin ) = s Mg
F = k Mg / (cos - k sin )
This is the force needed to keep the box in constant velocity.
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k > s where s is the critical value.
The force needed to keep the box at rest will be little more than F, say ƒ and is now s
ƒ = s Mg / (cos - s sin )
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We have
F cos = k (Mg + F sin ) and
ƒ cos = s (Mg + ƒ sin )
ƒ / F = (s / k) {(Mg + ƒ sin ) / (Mg + F sin )}
s / k = [ ƒ / F ] { (Mg + F sin ) / (Mg + ƒ sin ) }
s = k * (ƒ/ F ) * { (Mg + F sin ) / (Mg + ƒ sin ) }