Alarge crate filled with physics laboratory equipment must be moved up an incline onto a truck the crate is at rest on the incline. what can you say about the force of friction acting on the crate? a. the frictional force points up the inclineb. the frictional force points down the inclinec. the frictional force is zero

A. The frictional force points up the incline.

Explanation:

If the crate is at rest, this means that no net force is acting on it.

In absence of friction, we have two forces acting on the crate; the normal force (which prevents that the crate fall through the incline surface), which is always perpendicular to the surface and upward) and gravity force (always downward).

If no net force acting, the 2nd Newton's Law can be expressed as follows:

Fnet = m*a = 0

This is a  vector equation, which can be expressed as two algebraic equations, just decomposing both forces along two axes ,perpendicular each other.

As normal force (Fn) has no component  along the incline, we can choose as our axes, one parallel to the incline, and another perpendicular, as we will only need to decompose Fg in two perpendicular components.

If the net force is 0, both components must be zero also:

Fx (calling x-axis to the parallel to the incline) = 0

Along the incline. we have a component of the gravity force only, as follows:

Fx = m*g*sin θ = 0

In the perpendicular direction, we have:

Fy = N-m*g*cosθ = 0 ⇒ N = m*g*cos θ (choosing the upward direction as positive)

Returning to Fx, it is clear, that for any angle θ other than 0º, there must be another force, that added to this force, gives 0 as result.

This force can't be another than friction force, which always opposes to the relative movement between the surfaces in contact each other.

As the component of gravity force along the incline tries to accelerate the crate downwards the incline, friction force must point up the incline, i.e., A is the right choice.

v=ir = 3x15 =45 v