A 1,4-N crate is being pushed across a level floor at a constant speed by a force of 270 N at an angle of 20.0° below the horizontal.
(a) What is the coefficient of kinetic friction between the crate and the floor?
μk =
(b) If the 270-N force is instead pulling the block at an angle of 20.0° above the horizontal, as shown in the figure (b), what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in part (a).
_______m/s2
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a) This is a force balance problem: the crate is moving at constant speed, which is equivalent to saying that the acceleration of the crate is zero: sum of forces in the horizontal on the crate is equal to the product of the mass and acceleration, so the issue is merely identifying the horizontal forces on the crate.
There is a force pushing the crate and a frictional force which are equal and opposite: the pushing force is 270 N at 20 deg below the horizontal and so must be decomposed into horizontal and vertical components.
The horizontal component is magnitude multiplied by the cosine of 20 degrees (253.7 N), the vertical component is the product of the magnitude and the sine of 20 degrees (92.3 N). Note that the vertical component is a downward force.
The horizontal pushing force is resisted by an equivalent frictional force (Ff): Ff = 253.7 N. Ff is also equal to the normal force multiplied by the coefficient of friction. In this case, the normal force is the sum of the weight of the crate (14 N) and the downward force applied by pushing the crate (92.3 N):
Ff = 253.7 N = μk*(14 N + 92.3 N): μk=0.17
b) In this part, the forces no longer balance.
The horizontal force exerted on the crate by pulling is still 253.7 N, but the frictional force is reduced. The vertical component of the pulling force still has a value of 92.3 N, but the direction has changed to opposite that of the crate weight. μk has not changed, so,
Ff = μk*(14 N – 92.3 N) = 222.3 N
The net horizontal force on the crate is the difference in the horizontal loads, 253.7 N – 222.3 N = 31.4 N.
From F = ma, the mass of the crate may be determined: 14 N = m*g; g = 9.81 m/s2; m = 142.7 kg
Still using F=ms, the horizontal acceleration of the crate is a = F/m = 31.4 N / 142.7 kg = 0.22 m/s2