Question

A crate of mass 45.0 kg is being transported on the flatbed of a pickup truck. The coefficient of static friction between the crate and the truck's flatbed is 0.370,and the coefficient of kinetic friction is 0.210.

(a) The truck accelerates forward on level ground. What is the maximum acceleration the truck can have so that the crate does not slide relative to the truck'sflatbed?
3.63 m/s^2 which WebAssign says is correct.

(b) The truck barely exceeds this acceleration and then moves with constant acceleration, with the crate sliding along its bed. What is the acceleration of the craterelative to the ground?

No clue how to figure this out?

Answer 1

Given
Mass of crate, m = 45 kg
coefficient of static friction,µs =0.360
coefficient of kinetic friction,µk = 0.230
a)The net force on the truck is
F = µs m g
ma = µs m g
a = 0.360 *9.8 m/s^2
a =3.5 m/s^2
Thus, the acceleration of the truck is 3.5 m/s^2
b) The net force on the crate is
F = µk m g
m a = µk m g
a = 0.230 *9.8 m/s^2
a = 2.25 m/s^2
Thus the acceleration of the crate relative to ground is 2.25 m/s^2

Answer 2

Given
Mass of crate, m = 45 kg
coefficient of static friction,µs =0.360
coefficient of kinetic friction,µk = 0.230
a)The net force on the truck is
F = µs m g
ma = µs m g
a = 0.360 *9.8 m/s^2
a =3.5 m/s^2
Thus, the acceleration of the truck is 3.5 m/s^2
b) The net force on the crate is
F = µk m g
m a = µk m g
a = 0.230 *9.8 m/s^2
a = 2.25 m/s^2
Thus the acceleration of the crate relative to ground is 2.25 m/s^2

Answer 3

The crate will not move as long as the force exerted through the motion of the truck is less than the static friction between the crate and the flatbed.However, once this is overcome, the crate is no longer subject to static friction but to kinetic which is usually less than static. In other words it ismore difficult to get something moving than it is to keep it moving.

(a) You need to balance the force from static friction with that provided on the crate by the forward motion of the truck.

Us = coefficient of static friction
Fs = static friction = (normal force)*Us = m*g*Us

Force pulling the crate = mass * acceleration = m*a

Equate the forces.
m*a = m*g*Us
a = g*Us = 9.81*0.28 = 2.7468 m/s^2

So the acceleration of the truck is 2.7468 m/s^2

(b) We now need to use the coefficient of kinetic friction in the above equation and then take the difference with the forward force to get the netacceleration. I will use the fact that the crate is still up to a maximu acceleration of g*Us (see above).

Uk = coefficient of kinetic friction
Fk = kinetic friction = m*g*Uk
g*Uk = acceleration due to friction

net acceleration = forward minus friction = g*Us - g*Uk = g(Us - Uk)
a = 9.81(0.289 - 0.160) = 9.81*0.129
a = 1.1772 m/s^2

So the acceleration of the crate relative to the ground is 1.772 m/s^2 and it is in the same direction as the truck is moving.

Answer 4

When the tire doesn't slip, it means the point of contact with the ground is at rest with respect to the ground. So, static friction acts

Max acceleration = μs *g = 0.37*9.8 = 3.626 m/s^2

(b)When tire slips, it means the point of contact with the ground is in motion and so kinetic friction acts.

Constant acceleration = μk * g = 2.058 m/s^2

Answer 5

The maximum force that can be exerted on the block isµs FN = µs M g where FN is the normal force
M a = µs M g and a =µs g = .27 * 9.8 = 2.65m/s2
If the truck accelerates more than this the frictional forceis not great enough to accelerate the block likewise.
M a = µk M g and a =µk g = .17 * 9.8 = 1.67 m/s2
This is the only force on the block to provideacceleration relative to the ground

Answer 6

Given
Mass of crate, m = 45 kg
coefficient of static friction,µs =0.370
coefficient of kinetic friction,µk = 0.210
a)The net force on the truck is
F = µs m g
ma = µs m g
a = 0.370 *9.8 m/s^2
a = 3.62600 m/s^2

Thus, the acceleration of the truck is 3.626 m/s^2
b) The net force on the crate is
F = µk m g
m a = µk m g
a = 0.210 *9.8 m/s^2
a = 2.058 m/s^2

Thus the acceleration of the crate relative to ground is 2.058 m/s^2

Answer 7

Given
Mass of crate, m = 45 kg
coefficient of static friction,µs =0.360
coefficient of kinetic friction,µk = 0.230
a)The net force on the truck is
F = µs m g
ma = µs m g
a = 0.360 *9.8 m/s^2
a =3.5 m/s^2
Thus, the acceleration of the truck is 3.5 m/s^2
b) The net force on the crate is
F = µk m g
m a = µk m g
a = 0.230 *9.8 m/s^2
a = 2.25 m/s^2
Thus the acceleration of the crate relative to ground is 2.25 m/s^2