Offset Torque: Solving the 1000N Problem

In summary, the torque is the result of a cross product of two vectors, with the magnitude being equal to rFsin(theta) where theta is the angle between the two vectors. This means that only the components of the vectors that are perpendicular to each other are considered. In the given scenario, the torque can be calculated by taking 1000N times h, or by using the length of the green line and finding the perpendicular component of the 1000N force. Similarly, for the weight, the torque can be calculated by taking mg times x2, where x2 is the perpendicular component of the distance between the wheel and the center of mass.
  • #1
DarkBlitz
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Hello guys,

I am doing some basic design work and can't remember how offset torques work.

upload_2017-3-28_16-13-8.png


I am doing the sum of the torques should be equal to 0 in order to find the tipping point around either of the wheels.

My problem is with the 1000N applied at to top taking the pivot point around the left wheel.

Should I be using the component of the 1000N which is at right angles to the distance to the pivot point? (the green line) and take the green line as the distance? should I use 'h' as the distance to the pivot point with 1000N as the force? or a combination of both?

The same for m*g in the diagram. Should I assume that the centre of mass is where the weight is applied and take the perpendicular component?

Thanks very much for your help guys! Been a long time since I looked into this.
 
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  • #2
By definition, the torque is the result of a cross product of 2 vectors, i.e. ##\mathbf{T} = \mathbf{r} \times \mathbf{F}##. The magnitude of this torque (what you are looking for) is ##T = rF\sin\theta## where ##\theta## is the angle between the 2 vectors. This basically means that only the components of the vectors that are perpendicular to each other are considered.

You can see it as ##r_{\perp} F## or ##r F_{\perp}## as you wish, since both will give you the same answer: ##rF\sin\theta##.

The following is a representation of ##r F_{\perp}##:

512px-Torque%2C_position%2C_and_force.svg.png

The following is a representation of ##r_{\perp} F##:

télécharger.gif

So in your case, you take 1000 N times h ( or ##r_{\perp} F##). You could use ##r F_{\perp}## by using the length of the green line and by finding the perpendicular component of the 1000 N force (the orange line), it would give the same answer. It is just more complicated in this particular case.

Similarly, for the weight, you take mg times x2. Where x2 is the perpendicular component of the distance between the wheel and the center of mass.
 
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  • #3
Thanks a lot! cleared it up!
 

FAQ: Offset Torque: Solving the 1000N Problem

1. What is offset torque?

Offset torque is a force that is applied at a distance from the point of rotation, causing an object to rotate. It is often used to solve problems involving equilibrium or motion.

2. What is the 1000N problem?

The 1000N problem refers to a hypothetical scenario where an object weighing 1000N is suspended from a lever or pulley system, causing an imbalance of forces. This problem can be solved using offset torque.

3. How is offset torque calculated?

Offset torque is calculated by multiplying the applied force by the distance from the point of rotation. The unit for torque is Newton-meters (Nm).

4. How can offset torque be used to solve the 1000N problem?

By applying an offset torque in the opposite direction of the 1000N force, the system can reach equilibrium. This can be achieved by adjusting the distance between the point of rotation and the applied force.

5. What factors affect the offset torque required to solve the 1000N problem?

The offset torque required will depend on the weight of the object, the distance from the point of rotation, and the angle at which the offset force is applied. The direction of the applied force also plays a role in determining the required offset torque.

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