# Homework Help: Torque in a x y plane

1. Apr 19, 2010

### mybrohshi5

1. The problem statement, all variables and given/known data

A force F = 1.90 N i + 2.80 N j is applied at the point x = 2.50 m, y = 0.00 m. Find the torque about the origin.

2. Relevant equations

t = r*F

3. The attempt at a solution

I do not really understand multiplying vectors in this type of problem.

Can anyone give me a brief explanation of what to do for problems like this.

Thank you

2. Apr 19, 2010

What is the definition of the torque, given as a vector product? You have both vectors, so you can calculate the torque.

3. Apr 19, 2010

### mybrohshi5

my book says the definition of torque vector is just t = r x F

I get you can you the cross product for i and j but i am a little confused on how to tell the direction.

So for this problem for the magnitude of the torque about the origin would be

1.9(0) - 2.5(2.8) = -7

The direction for the torque for this is in the +z direction. Can you explain to my why that is? i thought it would be -z direction

thanks for the help :)

4. Apr 19, 2010

### mybrohshi5

I think i get it a little more now. i am having trouble understanding the next part though.

It asks for the torque about the point x = -1.80 m, y = 2.60 m?

so for my cross product i have (2.5i - (-1.8i + 2.6j) x (1.9i + 2.8j) which equals (4.3i)(2.8j) - (1.9i)(2.6) which therefore equals 7.1 but the answer is 17.0

17 would come from (4.3i)(2.8j) + (1.9i)(2.6)

i dont get why you would add them instead of subtract them?

5. Apr 19, 2010

### PhanthomJay

Whether the torque is plus or minus depends on the sign convention you use. If you take counterclockwise torques as negative, then the z axis is positive going into the plane of the page. If you take counterclockwise torques as positive, then the z axis is positive coming out from the plane of the page toward you. By convention, the z axis is usually considered positive coming out from the plane towards you, hence, countrerclockwise torques are positive (right hand rule).
Personally, I despise the cross product definition of torque, and use it only when absolutely necessary. It is often simpler to break the force into its x and y components, then the torque of each component force about the point in question is the component force times the perpendicular distance from the line of action of that force to that pivot point, then just algebraically add up the torques from each force component, ccw is +, and cw is - . In your example, both components produce ccw torques, so they are additive.

6. Apr 19, 2010

### mybrohshi5

Thanks Jay that helped so much. You just explained that 100 times better than my professor :)