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 10000kg load sits on the flatbed of a truck moving at 12m/s. Assume that the load is not tied down to the truck and assume that the coefficient of friction between the load and the truck bed is $0.5$ . Calculate the minimum stopping distance for which the load will not slide forward relative to the truck.

Answer 1

Hint: In this solution, we want to determine the minimum stopping distance such that the load does not slip and we can do this by calculating the distance corresponding to the maximum rate with which the truck can de-accelerate before the load starts slipping.
Formula used: In this solution, we will use the following formulae:
Third equation of kinematics: ${v^2} - {u^2} = 2ad$ where $d$ is the distance travelled by the truck with an initial velocity $u$, final velocity $v$, and acceleration $a$
${F_f} = \mu mg$ where ${F_f}$ is the friction force acting on a block of mass $m$ and coefficient of friction $\mu $ and gravitational acceleration $g$

Complete step by step answer:
When the truck is braking, there will be deceleration acting on the load. This deceleration should be such that the friction force acting on the load does not exceed the maximum friction force acting on the block. Since there is no vertical movement, the friction force will be
${F_f} = \mu mg$
The maximum deceleration of the truck will be corresponding to the maximum force that can be balanced by the friction force. So, if the acceleration of the truck is $a$,
$ma = \mu mg$
So, the maximum deceleration of the truck will be
$a = \mu g$
$ \Rightarrow a = 0.5 \times 10 = 5\,m/{s^2}$
Then if the truck is moving with 12m/s, it will require a stopping distance $(d)$ that is calculated by the third law of kinematics:
${v^2} - {u^2} = 2ad$
As the truck comes to a stop, its final velocity $v = 0$, $u = 12\,m/s$ and $a = 5\,m/{s^2}$, we get
$0 - 144 = 2( - 5)(d)$
Which gives us
$d = \dfrac{{144}}{{10}}$
$ \Rightarrow d = 14.4\,m$

Note: We’ve taken the acceleration in the above equation as negative as the truck will be decelerating. Here we have assumed that the truck can move with a constant deceleration such that the load does not slip on its bed.