Question

A duster weighs 0.5 N. It is pressed against a vertical board with a horizontal force of 11 N. If the coefficient of friction is 0.5, the minimum force that must be applied on the duster parallel to the board to move it upwards is
A. 0.4 N
B. 0.7 N
C. 6 N
D. 7 N

Answer 1

Hint: We can draw the free body diagram for the given duster. The gravitational force due to weight and the frictional force with the surface tend to resist the motion of the duster. The minimum force required to move the duster upwards is equal to the sum of these forces.

Complete step by step answer:

The following free body diagram shows a duster which has been pressed against a vertical board with a force of 11 N. This normal force decides the amount of the friction between the board and the duster. Therefore, we can write
$N = 11N$
If the coefficient of friction between the board and the duster is given as
$\mu = 0.5$
Then the force of friction between the board and the duster can be written as
$f = \mu N = 0.5 \times 11 = 5.5N$
Also the force due to weight of the duster acts along the direction of the frictional force and resists the motion of the duster. It value is given as
$mg = 0.5N$
The minimum force that must be applied in the upward direction in order to move the duster upwards along the board is signified as F. Based on the free body diagram, we can write the following equation for the forces.
$F = f + mg$
Inserting the known values, we obtain
$F = 5.5N + 0.5N = 6N$
This is the minimum force that must be applied on the duster parallel to the board to move it upwards.

Hence, the correct answer is option C.

Note:
It should be noted that when an object is placed horizontally on a horizontal surface, then the frictional force is due to the weight of the object which produces a normal force. In our case, we have pressed the duster on a vertical surface to produce a normal force which creates friction between the board and the duster.