A box of mass 5 kg rests on a horizontal table. The coefficient of static friction between the box and the table is 0.3. When the box is pulled with a horizontal force of 10 N, it does not move....

A box of mass 5 kg rests on a horizontal table. The coefficient of static friction between the box and the table is 0.3. When the box is pulled with a horizontal force of 10 N, it does not move. The force of friction exerted by the table on the box is
(1) 15 N
(2) 12 N
(3) 10 N
(4) 5 N
(5) 3 N

Asked on by saj-94

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valentin68's profile pic

valentin68 | College Teacher | (Level 3) Associate Educator

Posted on

Because the box is not moving the resultant force on it, on the horizontal and vertical axis need to be zero.

On the vertical the weight `G` of the box is balanced by the ground vertical reaction `N`. On the horizontal axis the applied force `F` need to be equal to the static friction force `Ff` . Thus by applying an increasing force to move the box, the static friction force will increase correspondingly until its maximum possible value

`mu*N =mu*G =mu*m*g =0.3*5*10 =15 N`

when the box will start moving.

Again if the box is not moving the applied force is equal to the friction force (otherwise there will be a resultant acceleration of the box).


Therefore the correct answer is 1) 15 N

Sources:
jeew-m's profile pic

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

When we pull the box with 10N force if the box is not moving that means the frictional force is equal to the applied force .S o the friction force will be 10 N at this moment.

When we increase the force friction force will also increase until its maximum.

Assume that `g = 10m/s^2`

Max friction `= 5gxx0.3 = 1.5g = 15N`

So the answer is 15N. It is at option (1).

Once the applied force exceed 15 N the box will start to move.

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