Question

To pull a box up a rough slope,the force required will be least when it is applied :
a) horizontaly
b) vertically
c) perpendicular to the plane
d) parallel to the plane

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Answer 1

Answer:

The correct option is;

d) Parallel to the plane

Explanation:

The forces acting on the box of mass, m are;

Weight of the box acting an angle, θ, equal to the inclination of the plane to the perpendicular of the plane

Weight of the box acting along the plane = m×g×sin(θ)

The force of friction along the plane = μ×m×g×cos(θ)

The total force acting downward along the plane $F_{down}$, = m×g×sin(θ) + μ×m×g×cos(θ)

The Force needed to pull the box up along the plane F = The total force acting downward along the plane

F = m×g×sin(θ) + μ×m×g×cos(θ) = m×g×(sin(θ) + μ×cos(θ))

When the force, Fₐ is applied vertically, the force acting along the plane = Fₐ×cos(θ)

When the force is applied perpendicular, the force acting along the plane = Fₐ×sin(θ)

When the force is applied horizontally, the force acting along the plane = Fₐ×cos(θ)

When the force is applied parallel to the plane, the force acting along the plane = Fₐ

Therefore, since Fₐ > Fₐ×cos(θ) and Fₐ > Fₐ×sin(θ), for acute angles, we have that the least force is required when the force is acting parallel to the plane.