Question

Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slide forward, crushing the cab, if the truck stops suddenly in an accident or even in braking. Assume, for example, that a 14,000-kg load sits on the flatbed of a 20,000-kg truck moving at 12.0 m/s. Assume the load is not tied down to the truck and has a coefficient of static friction of 0.400 with the truck bed.
(A) Calculate the minimum stopping distance for which the load will not slide forward relative to the truck.m
(B) Is any piece of data unnecessary for the solution? a) mass of the load
b) mass of the truck
c) velocity
d) coefficient of static friction
e) all are necessary

Answer 1

A) 18.4 m

B)

a) mass of the load

b) mass of the truck

Explanation:

A)

In order for the oad not to slide, its acceleration must be the same as the acceleration of the truck.

Since there is only one force acting on the load (the force of static friction), the acceleration of the load will be equal to the force of friction divided by the mass of the load (Newton's second law of motion):

$a=\frac{F_f}{m}=\frac{-\mu mg}{m}=-\mu g$ (1)

where

m is the mass of the load

$\mu=0.400$ is the coefficient of static friction

$g=9.8 m/s^2$ is the acceleration due to gravity

The acceleration of the truck (and the load) is also related to the stopping distance of the truck by the suvat equation:

$v^2-u^2=2as$ (2)

where

v = 0 is the final velocity of the car

u = 12.0 m/s is the initial velocity

a is the acceleration

s is the stopping distance

Since the acceleration must be the same, we can substitute (1) into (2), and solving for s we find:

$v^2-u^2=-2\mu g s\\s=\frac{v^2-u^2}{-2\mu g}=\frac{0^2-12.0^2}{-2(0.400)(9.8)}=18.4 m$

B)

From part A, we see that the data that we have not used in the calculation are:

- The mass of the load

- The mass of the truck

Therefore, the two pieces of data unnecessary for the solution are

a) mass of the load

b) mass of the truck