Question

A truck moving at 72kmph carries a steel girder which rests on its wooden floor. What is the minimum time in which the truck can come to stop without the girder moving forward? Coefficient of static friction between steel and wood is 0.5.

Answer 1

Given:

A truck moving at 72kmph carries a steel girder which rests on its wooden floor. Coefficient of static friction between steel and wood is 0.5.

To find:

Minimum time in which the truck can come to stop without the girder moving forward.

Calculation:

The girder will stay in translational equilibrium when the frictional force will be balanced by the the pseudo force when the truck is decelerating.

Initial velocity = 72 km/hr = 20 m/s.

Let mass of girder be m ;

$\therefore \: F = friction$

$= > ma = - \mu \: N$

$= > ma = \mu (mg)$

$= > a = - \mu g$

$= > \dfrac{0 - u}{t} = - \mu g$

$= > \dfrac{0 - 20}{t} = - (0.5) \times (10)$

$= > \dfrac{20}{t} = - 5$

$= > t = 4 \: sec$

So, final answer is:

$\boxed{ \sf{ \red{ \large{time = 4 \: sec}}}}$

Answer 2

F.B.D. of Girder/Block :-

$\setlength{\unitlength}{1cm}\begin{picture}(6,6)</p><p>\put(3, 2){\line(0, 1){1}}</p><p>\put(3, 2){\line(1, 0){1}}</p><p>\put(4, 3){\line(0, -1){1}}</p><p>\put(3, 3){\line(1, 0){1}}</p><p>\put(3.5,2.5){\circle*{0.09}}</p><p>\put(3.5,2.5){\vector(0,1){1}}</p><p>\put(3.5,2.5){\vector(0,-1){1}}</p><p>\put(3.5,2.5){\vector(1,0){1}}</p><p>\put(3.5,2.5){\vector(-1,0){1}}</p><p>\put(3.4,3.59){\sf{N}}</p><p>\put(3.3,1.29){\sf{mg}}</p><p>\put(1,2.4){\sf{P.F. = ma}}</p><p>\put(4.6,2.4){\sf{friction}}</p><p>\end{picture}$

$\sf{\\}$

We will solve using Non Inertial Frame of Reference. So there will be a Pseudo Force = ma which you can observe in the FBD

$\sf{\\}$

We can observe that:-

$\sf{N = mg \dots(i)}$

$\sf{\\}$

and,

$\sf{ma = friction}$

$\longrightarrow \sf{ma = {\mu}_{s}N}$

$\longrightarrow \sf{\cancel{m}a = {\mu}_{s}\cancel{m}g \quad \quad[From\:(i)]}$

$\longrightarrow \sf{{a}_{max} = {\mu}_{s}g}$

$\sf{\\}$

So, we have, v = 0m/s, $\sf{{a}_{max} = {\mu}_{s}g\:m/{s}^{2}}$, u = 72kmph = 20m/s

$\sf{\\}$

So,

$\sf{v = u - at \quad \quad (As\:it\:is\:retarding)}$

$\longrightarrow \sf{0 = 20 - {\mu}_{s}gt}$

$\longrightarrow \sf{20 = {\mu}_{s}gt}$

$\longrightarrow \sf{20 = 0.5 * 10 * t}$

$\longrightarrow \sf{t = 4s}$

As the acceleration is maximum $\sf{({a}_{max})}$, correspondingly, time will be minimum $\sf{({t}_{min})}$

$\sf{\\}$

$\longrightarrow \underline{\underline{\sf{{t}_{min} = 4\:seconds}}}$