Answer to Question #159665 in Classical Mechanics for Any

Question

Linear motion in an accelerating coordinate system

A box of mass m is placed at the centre of a flat truck bed while the truck is moving at

constant velocity V⃗ O . The truck bed has a length of L and its surface has a sliding friction

coefficient μ. The truck then accelerates with a constant acceleration

⃗

A

O for a few seconds.

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 and accidentally opens up its back

doors. (assume the box as a particle and g = 10 ms-2)

a)

Describe your system in detail and sketch the free-body diagrams (FBD) of the box

when it is sliding and falling off the truck.

b)

Write the equation of motion (EOM) of the box and its general solutions inside and

outside of the truck.

c)

Use relevant octave script provided to solve and plot the velocity and position of the

box as a function of time.

d)

Describe in detail the motion of the box based on the plots in part c)

Answer 1

Given ,

a)

mass of the block = m,

it is moving the the constant velocity $V_o$

When it is was moving the the constant velocity, then at then at that time there was only two force were working one is weight of the block due the gravity and other was the normal reaction of the block by the truck floor. but when the truck starts to accelerate with the acceleration $A_o$ then due to the pseudo force, due to which block try to slide over the floor of the truck, but due to the friction force it will not fall. Friction force will stop to fall the block on the ground but if the acceleration of the truck will greater than the friction force then it will fall on the ground. The below diagram shows the FBD of the system.

b)

c) Now, applying the force balance equation,

Let the resultant acceleration of the block be a.

$\Rightarrow mg = N...(i)$ (vertical force )

Horizontal force,

$mA_o-\mu N=ma...(ii)$

From equation (i) and (ii)

$a=A_o-\mu g$

d)$a= \frac{dv}{dt}$

$a=\frac{dv}{dx}\times \frac{dx}{dt}$

$\Rightarrow a = v \frac{dv}{dx}$

$\Rightarrow a dx = v dv$

$\Rightarrow (A_o-\mu g)dx=v dv$

Now, taking the integration of both side of the given equation,

$\int_{v_o}^vvdv=\int_{L/2}^L(A_o-\mu g)dx$

$\Rightarrow [\frac{v^2}{2}]_{v_o}^v=(A_o-\mu g)[L-\frac{L}{2}]$

$\Rightarrow v^2-v_o^2=2(A_o-\mu g)\frac{L}{2}$

$\Rightarrow v^2=v_o^2+(A_o-\mu g)L$