A 4111.0 kg truck is parked on a 14.6° slope. What is the friction force on the truck?
see other post.
one of the links have the answer
q.gs/D2Tva
q.gs/D2Tvb
q.gs/D2Tvc
q.gs/D2Tvd
q.gs/D2Tve
q.gs/D2Tvf
q.gs/D2Tvg
To find the friction force on the truck, we need to consider the forces acting on it. The relevant forces in this case are the gravitational force (mg) acting vertically downward and the component of the gravitational force parallel to the slope (mg*sinθ) acting downhill.
First, we need to find the component of the gravitational force parallel to the slope. The formula to find this component is given by:
Parallel Force (Fparallel) = mg * sinθ
where m is the mass of the truck and θ is the angle of the slope.
In this case, the mass of the truck is given as 4111.0 kg and the angle of the slope is 14.6°. So we can substitute these values into the formula:
Fparallel = 4111.0 kg * 9.8 m/s^2 * sin(14.6°)
Calculating this, we get:
Fparallel = 4111.0 kg * 9.8 m/s^2 * 0.2508
Fparallel ≈ 10134.3 N (rounded to the nearest tenth)
Now, the friction force opposes the motion of the truck and acts uphill to balance out the downhill force acting parallel to the slope. Therefore, the friction force will have the same magnitude as the parallel force but in the opposite direction.
So, the friction force on the truck is approximately 10134.3 N, acting uphill.