Question

A box is loaded on open horizontal floor of truck.The coefficient of friction between the box and floor is 0.6 . The maximum magnitude of acceleration of truck so that the box does not slip back is (g=10m/s^(2))​

Answer 1

Answer:

Maximum frictional force acting =μmg=0.6×1×10=6N

Pseudo force acting on the body =ma=1×5=5N

∵ pseudo force < frictional force

The frictional force acting is only 5N.

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Answer 2

Concept:

Friction is the force that prevents solid surfaces, fluid layers, and material constituents from sliding past one another.

Given:

The mass of the box is 1kg. The coefficient of friction between the box and the floor is 0.6.

Find:

The maximum magnitude of the acceleration of the truck so that the box does not slip.

Solution:

The coefficient of friction, $\mu=0.6$

The mass of the box, $m=1kg$

The maximum frictional force acting on the body is given as:

f = μmg = 0.6 × 1 × 10 = 6 N

Now, the pseudo force acting on the body is:

$F=ma=f\leq \mu N=\mu mg$

Therefore,

$ma_{max}=\mu mg\\a_{max}=\mu g=0.6\times10=6m/s^2$

Therefore, the maximum magnitude of the acceleration of the truck so that the box doe snot slip back is $6m/s^2$.

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