Answer to Question #159736 in Classical Mechanics for Nurulain Fatihah

Question

1.

A small box of mass m is located at the middle point of the front end of a flat truck bed while the 

truck is stopped at a traffic light. The truck bed has a length L = 2.2 m , a height H = 1.2 m off 

the the road surface and its surface has a sliding friction coefficient μ = 0.05. The truck then 

accelerates with a constant acceleration AO = 1.1ms

−2

for a few seconds the traffic light turns 

green. Assume the origin O’ of the moving coordinate system coincides with the centre of the 

box and it is aligned perfectly vertically with the origin O of the static coordinate system on the 

road at the exact instant when the truck starts to accelerate. (assume the box as a particle and g = 

10 ms-2)

a)

Assume that both the x-axis and x’-axis for both coordinate systems are parallel to the 

length of the truck bed and the motion is observed by a non-inertial observer (NIO) in the 

truck and by an inertial observer (IO) on the roadside.

i)

Sketch the free-body diagrams (FBDs) of the box when it starts sliding according to 

both the NIO and the IO.

ii)

Write the equations of motion (EOMs) of the box and their general solutions.

b)

Use the octave script provided to solve and plot the velocity and position of the box as a 

function of time.

i)

Describe the motion of the box according to the NIO and the IO.

ii)

What are the accelerations of the box according to the two observers?

iii)

Determine the velocities of the box when it hit the back door.

iv)

According to the IO what is the position of the box when it hits the back door?

Answer 1

Given,

Mass of the box=m

Length of the truck bed (L) = 2.2m

Height of the truck bed (H)=1.2m

Coefficient of friction $(\mu)=0.05$

Acceleration of the truck $(a_{o})=1.1 m/s^2$

a (i)Here X-Y plane is a inertial frame and x'-y' is non inertial frame. When truck starts to accelerate, then due to friction force it will not slide for some time, but as we increase the acceleration, then pseudo force on the block will get increase, and when the pseudo force become the greater than the friction force then block starts to slide.

a(ii) $f_s= \mu N =0.05\times m\times10=0.5mN$

pseudo force on the block $f= ma= 1.1mN$

from the above we can see that pseudo force on the block is greater than the friction force so it will not slide. But with respect to inertial force it will move. Velocity of the block with respect to inertial frame $(v)= 1.1t$

and distance covered by the block with respect to inertial frame $d= \frac{1.1t^2}{2}$

b) t = [-100:5:100];
d = 0.555*t.^2;
plot(t, d)

$v^2=2\times 1.1\times s$

$x=\frac{v^2}{2.2}$

Octave script-

x = [-100:5:100];
x = 0.4545*y.^2;
plot(x, y)