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

Expert Answers
valentin68 eNotes educator| Certified Educator

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

jeew-m eNotes educator| Certified Educator

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.