# Is there a way to calculate an offset force created by torque?

I am trying to figure out how to calculate the force created by an offset torque. The theoretical scenario is a 2x4 piece of lumber mounted by a hinge on the wall at exactly 90°. The weight of the 2x4 creates torque on the hinge, making the hinge want to swing down, but there is a wall in the way preventing the hinge from swinging: the wall receives force created by the torque, in this case offset by 4 inches (assume a 2x4 is literally 2"x4").

I think I can calculate the torque (measured at the hinge) correctly using torque = force * distance, because the force is the weight of the lumber in foot-pounds, and the distance is x/2 (since the weight is distributed evenly over the whole length x, we can pretend all of the weight is halfway).

So with w = weight in foot-pounds x = length in feet the equation is:

torque = w * x/2

So if the board is 2 foot, the torque is equal to the weight. If the board is 4 feet, the torque is double the weight, etc.

But how do you translate that torque into the force that is applied at the bottom of the 2x4?

Warning: I'm about to try it below, but if you already know the correct way to do it, feel free to skip the rest of this and just provide the correct answer!

I know force = torque / distance so if d = distance from hinge to pressure point in feet

force = (w*(x/2)) / d

So in this example the distance is d = 4 inches = 1/3 feet, we can graph it:

For a 6' board, the force would be 6 times the weight.
For an 8' board, the force would be 12 times the weight.
For a 20' board, the force would be 30 times the weight.

Is this the right approach?

• Graphs made with desmos.com Jul 5, 2021 at 17:35
• Ok, why do people keep assuming that this is homework?? If I was in a class it would be so much easier to get the answer from a textbook than to create this question. This is a simplified version of the question I asked here last year. Jul 6, 2021 at 14:45

You can think it this way. There are 2 forces with non zero torques:

1. The weight force $$W$$ acting on the centre of gravity
2. Reaction force $$R$$ from the cube acting horizontally

Since the board is in rest, those torques should compensate each other. Torque is the force (red) times arm of the forces (green). We have an equation:

$$W\cdot \frac x2 = R\cdot d.$$

From this you can express $$R=W\frac{x}{2d}$$

The normal force from the wall acting on your “cube” must produce a CCW torque about the hinge which matches the CW torque produced by the weight of your board.

• Would that mean that the force is always equal to the torque no matter how far the cube is from the hinge? Jul 5, 2021 at 19:56
• Moving the cube closer to the hinge would require a larger force to produce the same torque. ( Force can only be numerically equal to torque if the radius is one) Jul 6, 2021 at 14:16
• I guess I am having a hard time with this because I am trying to find that expressed not as torque but a linear force. The cube is being crushed on both sides, as if in a press. I'm trying to find the force of the press. Jul 6, 2021 at 14:59