Question

A body is under the action of two forces 7N and 10N. Find the resultant of the two forces if: 1.) The forces are parallel and act in the same direction
2.) The forces are parallel and act in opposite directions.
3.) The forces are inclined at an angle of 60° to each other.
4.) The forces are inclined at an angle of 160°.
5.) The forces act at 90° to each other.

Answer 1

Answer:

1) 17 N

2) -3 N

3) 14.80 N

4) 4.18 N

5) 12.21 N

Explanation:

  The resultant (R) of two forces P and Q acting at an angle θ is given by

           $R=\sqrt{P^2+Q^2+2PQcos\theta }$

 We have P = 7N  and Q =10N

 1) The forces are parallel and act in the same direction ( θ = 0° )

       $R=\sqrt{7^2+10^2+2*7*10*cos0}=\sqrt{(7+10)^2)}=17N$

2) The forces are parallel and act in opposite directions( θ = 180° )

    $R=\sqrt{7^2+10^2+2*7*10*cos180}=\sqrt{(7-10)^2)}=-3N$

3) The forces are inclined at an angle of 60° to each other( θ = 60° )

    $R=\sqrt{7^2+10^2+2*7*10*cos60}=14.80N$

4) The forces are inclined at an angle of 160°( θ = 160° )

    $R=\sqrt{7^2+10^2+2*7*10*cos160}=4.18N$

5) The forces act at 90° to each other( θ = 90° )

    $R=\sqrt{7^2+10^2+2*7*10*cos90}=12.21N$