Question

a block of mass 10kg is placed on rough surface the coefficient of static friction between block and surface is 0.18 what force parallel to surface is needed to just began its motion.​

Answer 1

Answer:

the answer is 30kg OK

Explanation:

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Answer 2

Given:

Mass of the block $=10kg$

Coefficient of static friction $=0.18$

To find:

Force required to just begin the motion of the box.

Solution:

If we see the free body diagram of the 10 kg mass, we can see that there are 2 forces acting on the box.

$1)$ The force of gravity which is equal to the weight of the block.  

                               $W=mg$

Since, the block is stationary on the ground and not penetrating in the ground. Therefore a force similar and opposite to the direction of force of gravity is being applied by the box on the ground. This will be the normal reaction of the block of mass on the ground.

Hence,                

$R-mg=0$  

$R=mg$              $(i)$

$2)$ The external force $(F_{ext} )$ applied on the box so that the box could just move on the ground.  

We know, initially the box will not move because a force acting in the direction opposite to the direction of the application of external force.

This force is the frictional force $(F_{s} )$

$F_{s}=$ μ$_{s} R$            $(ii)$

Where, μ$_{s}$ is the coefficient of static friction.

To overcome this force of friction, a force greater than or equal to the force of friction should be applied so that the box starts moving on the ground.

$F_{ext}\geq F_{s}$

$F_{ext}\geq$ μ$_{s}R$

$F_{ext}\geq$ μ$_{s}mg$           From $(i)$

$F_{ext}\geq 0.18$ × $10$ × $10$

$F_{ext}\geq 18N$

Final answer:

Hence, application of any force greater than 18 N will result in the motion of the body.