Answer:
a) The truck must have an acceleration of at least 3.92 * cos(θ)^2 for the stack to start sliding.
b) a = 2.94 * cos(θ)^2 + 0.32 * L * cos(θ)
Explanation:
The stack of plywood has a certain mass. The weight will depend on that mass.
w = m * g
There will be a normal reaction between the stack and the bed of the truck, this will be:
nr = -m * g * cos(θ)
Being θ the tilt angle of the bed.
The static friction force will be:
ffs = - m * g * cos(θ) * fJK
The dynamic friction force will be:
ffd = - m * g * cos(θ) * fik
These forces would produce accelerations
affs = -g * cos(θ) * fJK
affd = -g * cos(θ) * fik
affs = -9.81 * cos(θ) * 0.4 = -3.92 * cos(θ)
affd = -9.81 * cos(θ) * 0.3 = -2.94 * cos(θ)
These accelerations oppose to movement and must be overcome by another acceleration to move the stack.
The acceleration of the truck is horizontal, the horizontal component of these friction forces is:
affs = -3.92 * cos(θ)^2
affd = -2.94 * cos(θ)^2
The truck must have an acceleration of at least 3.92 * cos(θ)^2 for the stack to start sliding.
Assuming the bed has a lenght L.
The horizontal movement will be over a distance cos(θ) * L because L is tilted.
Movement under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
In this case X0 = 0, V0 = 0, and a will be the sum of the friction force minus the acceleration of the truck.
If we set the frame of reference with the origin on the initial position of the stack and the positive X axis pointing backwards, the acceleration of the truck will now be negative and the dynamic friction acceleration positive.
L * cos(θ) = 1/2 * (2.94 * cos(θ)^2 - a) * 0.4^2
2.94 * cos(θ)^2 - a = 2 * 0.16 * L * cos(θ)
a = 2.94 * cos(θ)^2 + 0.32 * L * cos(θ)
