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# Physics problem I don't understand?

A 3.0 m rod is pivoted about its left end. A force of 6.0 N is applied perpendicular to the rod at a distance of 1.2 m from the pivot causing a counter clock wise torque, and a force of 5.2 N is applied at the end of the rod 3.0 m from the pivot. The 5.2 N is at an angle of 30 degrees to the rod and causes clockwise torque. What is the net torque about the pivot?

*Please help! this problem is on my review sheet for a physics test I have tomorrow and I don't understand it! I know that the answer is -0.6 N*m but I don't know how to get it.

### 2 Answers

- KododoLv 41 decade agoFavorite Answer
....

The counterclock torque of the 6.0N force is 6.0(1.2)=7.2N*m

the 5.2N force which is at a 30degree down, creating a clockwise torque at the end of the 3m rod torque has 2 resultant forces acting on the rod.....a downward force creating the clockwise torque, and an outward force inline with the rod PULLING against the pivot.

that force form a triangle with an included angleof 30*

The rotational force is 5.2sin30=2.6N its torque is 2.6N(3)=7.8N*mclockwise

The RESULTANT torque is 6.0(1.2)-2.6(3)=7.2-7.8= -.6N*m clockwise

NOTE.....the force at an angle has two vectors...one in line with the rod and one downward.

Since you have a review coming be aware, of a few unique set up:::::you will be given forces pulling against each other, either in line, or at angles to each other....keep in mind.. convert the at angle- forces into their components, then just add or subtract the component forces to get a final result.

good luck...

- Anonymous1 decade ago
Because the 5.2 force is at a 30 degree angle it gives a component force perpendicular to the rod of 5.2sin30 = 2.6.

So the net torque is 6x1.2 - 2.6x3 = -.6