498,965 views
21 votes
21 votes
A force of 30.ON is acted on a block of mass 15.0kg placed on a horizontal surface but the boxdoes not move. Calculate the magnitude of the frictional force between the surfaces when theapplied force is along the horizontal and when it makes angle 60° with the horizontal.

User Nathan Getachew
by
2.9k points

1 Answer

13 votes
13 votes

We will have the following:

1. In the first case, we will calculate the normal force:


N=(15\operatorname{kg}\cdot9.8m/s^2)\Rightarrow N=147N

Then, we will have that the fricion coefficient will be:


\mu_f=(30.0N)/(147N)\Rightarrow\mu_f=(10)/(49)\Rightarrow\mu_f\approx0.20

So the magnitude of the friccional force when applied horizontally will be of 30.0N.

2. And the magnitude when applied horizontal but the block is on an inclined surface of 60°, we will have that:


F=(10/49)(15.0\operatorname{kg}\cdot9.8m/s^2)\cos (60)\Rightarrow F=15N

So, the friction force when the force is applied, is 15N.

User Sangcheol Choi
by
3.2k points