The answers entered were incorrect. please help. Problem 5.34 One straightforwar

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The answers entered were incorrect. please help.

Problem 5.34 One straightforward way to measure the coefficients of friction between a box and a wooden surface is illustrated in the figure. The sheet of wood can be raised by pivoting it about one edge. It is first raised to an angle 1 (which is measured) for which the box just begins to slide downward. The sheet is then immediately lowered to an angle 2 (which is also measured) for which the box slides with constant speed down the sheet(Figure-1) Figure 1 of 1 8 Plywood sheet

Explanation / Answer

The normal force on the block, N = mgcos()

Analyse the forces parallel to the slope. Take uphill as positive; downhill as negative.

The component of the box's weight parallel to the is slope is P = -mgsin()

If the block is on the point of slipping =. If s is coefficient of static friction, then the limiting frictional force is L= sN = s mgcos()

The resultant force in the direction parallel to the slope is F = L + P

Acceleration is zero, so applying F = ma (from Newton's 2nd law) in the direction parallel to the slope:
F = ma
L + P = m*0 (since the resultant force is L+P
s mgcos() + ( -mgsin()) = 0
s cos() - sin() = 0
s = sin()/cos()
s = tan()

b)
To find the coefficient of kinetic friction, repeat the above working using k instead of s, and using instead of .

The working is exactly the same, giving:
k = tan()

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