# Calculating the net torque help

• cece
In summary, the net torque about the pivot point is -28 Nm, with a vertical force of 6 N acting at a distance of 2 m and a vertical force of -8 N acting at a distance of 5 m. The horizontal force of 10.39 N does not contribute to the torque.
cece
Calculate the net torque about point for the two forces applied as in the figure below

The rod and both forces are in the plane of the page.

torque= ? Nm

http://session.masteringphysics.com/problemAsset/1026433/5/yg.10.42.jpg

i know that F2 = 0 since its pointing through point 0, but for F1 I've so far got 16, 0, -16, -24... which were all wrong. I'm using the torque=F x L... which includes sin theta.

F2 is acting on the rod at an angle, therefore there is a componant of the force acting through the pivot, and a componant acting vertically upwards.

So first things first, you need to determine the upward componant of F2.

Then you will have two forces acting on the rod at two distances, from which you can calculate the overall torque.

ok. am i on the right track?
12x3xsin30=18
8x2xsin90=16
18+16=34Nm?

Well, you're almost there. I think you flipped the distances round, 12*3 and 8*2. However, F1 is acting at a distance of 5 m (2 + 3) from the pivot.

Also, you're adding the forces, but they aren't acting in the same direction.

I think it would help you to follow things through if you were to resolve the F2 force to find it's vertical and horizontal componants, then re-draw the diagram given to you. Once you have this picture in front of you, you will find it a lot easier to see what's going on.

i don't know how to find the x and y components to tell you the truth.

so should my process be:
12x2xsin30=18
8x3xsin30=-16
18-16=2Nm?

OK, forget F1, F1 is fine as it is because it's only acting vertically, there is no horizontal componant to it.

For F2 though, you just did find the vertical componant, that's what 12sin30 is. That is the force acting upward at a distance of 2m.

Now bring back F1, which is a force of 8N acting at a distance of five metres from the pivot.

If we assume the up direction as positive, that means we have a force of 12sin30 at 2m acting up, and a force of -8N at 5m acting up.

oooh so it would be

12x2xsin30=18
8x5sin30=20

18+20=38

(i'm sorry I am really not good at this)

OK, crash course in resolving forces.

You have 8N force acting at a 90 degree angle to the horizontal bar. Therefore the vertical componant of that force is 8 * sin90 = 8*1 = 8N. The horizontal componant of that force is 8 * cos90 = 8*0 = 0N. Therefore the full 8N force is acting vertically downwards on that bar.

You also have a 12N force acting on the bar at a 30 degree angle. That's 12 * sin30 = 6 N vertically and 12 * cos30 = 10.39 N acting horizontally. Since the horizontal force acts through the pivot, it doesn't contribute anything to the torque, so we can ignore it.

So now the situation is like this:
Code:
   6 N
|
|
----|------|
|
|
8 N
Taking the upward direction as positive we can show the diagram like this:
Code:
   6 N    -8 N
|      |
|      |
----|------|
Now try calculating the torque from that last diagram.

-8N x 5m = -40 Nm
6N x 2m = 12 Nm
-40 + 12 = -28Nm

## What is torque?

Torque is a measure of the force that causes an object to rotate around an axis. It is calculated by multiplying the force applied to an object by the distance from the axis of rotation to the point where the force is applied.

## What is net torque?

Net torque is the sum of all the individual torques acting on an object. It takes into account both the direction and magnitude of each torque.

## How do you calculate net torque?

To calculate net torque, you need to determine the direction and magnitude of each individual torque acting on an object. Then, you can use the equation net torque = Σ (torque) to find the total torque.

## What is the unit of torque?

The unit of torque is typically measured in newton-meters (Nm) in the metric system or foot-pounds (ft-lb) in the imperial system.

## Why is calculating net torque important?

Calculating net torque is important because it helps us understand how an object will rotate and whether it will be in equilibrium or not. It is also essential in many engineering and scientific applications, such as designing machines and analyzing the motion of objects.

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