A box sits on a horizontal wooden ramp. The coefficient ofstatic friction between the box and the ramp is 0.30. You grab oneend of the ramp and lift it up, keeping the other end of the rampon the ground. What is the angle between the ramp and thehorizontal direction when the box begins to slide down the ramp?
May I know θ = tan^-1 (μs) is the general equation for this question or it differs from other questions,and we would have to derive another equation for a different question?
May I know θ = tan^-1 (μs) is the general equation for this question or it differs from other questions,and we would have to derive another equation for a different question?
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Your equation is correct for the situation.
The normal reaction force is mg cos (theta)
The maximum friction is μs mg cos( theta)
The sliding force on the object is mg sin( theta)
Equate the two
μs mg cos( theta)= mg sin (theta)
so
μs =sin(theta) / cos( theta) = tan(theta)
therefore
theta = atan(μs)
Just as you state in your question
The normal reaction force is mg cos (theta)
The maximum friction is μs mg cos( theta)
The sliding force on the object is mg sin( theta)
Equate the two
μs mg cos( theta)= mg sin (theta)
so
μs =sin(theta) / cos( theta) = tan(theta)
therefore
theta = atan(μs)
Just as you state in your question