What force is required to push a block (mass m) up an inclined plane that makes an angle of θ with the horizon at a constant vel
ocity, if the coefficient of friction between the plane and the block is μ? Group of answer choices
a. mg μ cosθ
b. mg (μ cosθ + sinθ)
c. mg (μ sinθ + cosθ)
d. mg (μ cosθ + μ sinθ)
e. mg (μ cosθ + m sinθ)
a. mg μ cosθ
b. mg (μ cosθ + sinθ)
c. mg (μ sinθ + cosθ)
d. mg (μ cosθ + μ sinθ)
e. mg (μ cosθ + m sinθ)
1 answer:
Answer:
b. mg ( μ · cos θ + sin θ)
Explanation:
Hi there!
Please, see the attached figure for a graphical description of the problem.
We have the following parallel forces acting on the block (in parallel direction to the direction of movement):
F = applied force.
Fr = friction force.
wx = parallel component of the weight.
According to Newton´s second law:
∑F = m · a
Where "m" is the mass of the block and "a" its acceleration.
Then:
F - Fr - wx = m · a
Since the block is to be pushed at a constant velocity, the acceleration is zero. Then:
F - Fr - wx = 0
F = Fr + wx
The applied force has to be equal to the friction force plus the parallel component of the weight to push the block at constant velocity.
The friction force is calculated as follows:
Fr = μ · N
Where N is the normal force and μ is the coefficient of friction.
Notice that the normal force is of the same magnitude as the perpendicular component of the weight, wy.
Let´s apply Newton´s second law in the perpendicular direction to show this:
∑F = m · a
N - wy = m · a
The acceleration of the block in the perpendicular direction is zero. Then:
N - wy = 0
N = wy
And wy can be obtained by trigonometry (see figure):
wy = W · cos θ
N = wy = mg · cos θ
The parallel component of the weight is calculated using trigonometry (see figure):
wx = W · sin θ
wx = mg · sin θ
Then the applied force will be:
F = Fr + wx
F = μ · N + mg · sin θ (N = wy = mg · cos θ)
F = μ · mg · cos θ + mg · sin θ
F = mg ( μ · cos θ + sin θ)
The correct answer is the b.
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