- #1

blackice552

- 12

- 0

I have 2 problems:

1. We have a stick of length 30cm. The stick has a pivot point in its left end. We have a force 3.3N acting at angle 0.7rads and a distance 16.1cm from the pivot. What is the MAGNITUDE of the force we must apply perpendicular to the stick at a distance 5.0cm from the pivot so that the stick does NOT rotate?

2.I am applying a force 2.4N on a wrench at an angle 0.5rads from the horizontal. If the point of application of the force is 7.8cm from the pivot, what is the MAGNITUDE of the torque about the pivot?

2. Homework Equations :

Torque = Force * Perpendicular Distance

Perpendicualr Distance = distance * sin(theta)

Note: I'm not sure if it matters in problem one that the distances aren't the end of the object.

3. The Attempt at a Solution :

1.

Torque1 +Torque2 = 0

Torque1 = .161 m * (3.3 N) * (sin (.7)) = 3.422

Torque2 = -3.422 = F(.5)(sin (pi/2)) = .5F

-6.844 = F

Mag. F = 6.844 N

2.Perpendicular Distance = .078 m * sin(.5) = .037

Torque = 2.4 N * .037 m= .09 Nm