How to calculate lever force when lever has uniformed distributed load?

Question

We have a simple class 1 lever:

$$\begin {array}{c} \text {5,000 kg} \\ \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \\ ==================== \\ \hphantom {==} \triangle \hphantom {==============} \\ \vdash \text{1 m} \dashv \vdash \hphantom {======} \text{4 m} \hphantom {======} \dashv \\ \end {array} $$

The lever ($===$) is 5 m long. The fulcrum ($\triangle$) is 1 m from one end of the lever. The lever has a object sitting uniformly upon it weighing 5,000 kg.

How do I compute the upward force that needs to be exerted at the end of the 1 m side of the lever to keep the lever stationary? $F = (W \times X)/L$ is simple when the weight is applied at the very end of the lever. But what happens if the weight is distributed along the lever?

Our final goal is to tether the free end (on the 1m side) to keep the lever level and we need to know how strong the tether should be.

Answer 1

In any continuous situation, you simply use integration. The linear mass density of your block is $\lambda=\frac{m}{\ell}=$1000 kg/m. Now you can express the torque due to an infinitesimal slice of the rod of width $dx$ at position $x$ as $$ d\tau=(\lambda dx) * x * g $$ where $x$ is measured from the fulcrum. Finally, you just sum up all of the little torques from each infinitesimal slice with integration. $$ \tau = \lambda g\int_{-1}^{4}x\ dx=7.5\ g\lambda=73.5\ \text{kN*m} $$

Answer 2

Since the mass is 5k kg and the lever is 5m, this makes it quite easy to simplify because it is exactly 1k kg per m.

The leftmost 2k kg (2m) of the mass has its center of mass exactly above the fulcrum so can be ignored as it provides no contribution to the moment. This leaves 3k kg (3m) spread from 1m to 4m on the right side. The center of mass will therefore be at 2.5m.

Now it's super-simple, assuming you want the moment when the lever is level (i.e. when gravity is pulling straight down, perpendicular to the lever):

$$ \text{torque} = rF = rmg $$

• $ r $ is the radius (distance) in m (2.5).
• $ m $ is the mass in kg (3000).
• $ g $ is the acceleration due to gravity in $ \text{ms}^{-2} $ (9.80665).

$$ \text{torque} = 2.5 * 3000 * 9.80665 = 73549.875 \text{ Nm} $$

Since your edit/update indicates that you are looking for the upwards force at the 1m end, this will be the torque (from above) divided by the distance (1m). Which is therefore 73549.875 N.

Answer 3

To answer the new question, which it really rather different to the original question, you will require a 7500 g N downward force at the left-hand tip to balance forces.

Taking moments about your support (which is now, indeed, a pivot):

$$F_{\text{LHS free end}} * 1 = 5000*g * 1.5$$

$$F_{\text{LHS free end}} = 7500*g \text{ N}$$

In other words, yes, you can treat your distributed load as a point load acting at the centre of the beam. You can prove this my solving this by integration of the distributed load.

Answer 4

The effect of any bit of force along a lever is proportional to its distance from the fulcrum. This nice linear relationship works out so that for a rigid mass, you can simply model it as a point mass at its center of mass.

For weight effects (force due to mass and gravity), it's purely the horizontal distance from the fulcrum to the center of mass that matters. If you define X to the right and Y up in your diagram, then the Y coordinate of the mass is irrelevant. Note, however, that when the lever moves, the X coordinate of the mass also moves, especially when it is not right on the lever arm. For small movements of the lever, you my be able to ignore this.

Put more mathematically, the torque on the fulcrum is the vector from the fulcrum to the center of mass, cross the gravitational force on that mass. Since the latter is always down (-Y) in this example, only the X component of the vector to the mass matters in getting the toque magnitude.

Answer 5

A uniformly distributed load can be considered to act in its centre. Working in kg and m:

Clockwise moment about the left hand end = 5000 * 2.5 = 12500 Anticlockwise moment about the left hand end = F * 1 (where F is the reaction at the fulcrum)

These must be equal for it to be balanced, giving F=12500kg

Resolving vertically (the total downward force must equal the total upward force), taking T as the reaction on the tether: T + 5000 = 12500, therefore T = 7500kg.

Or converting into N (as you say you want a force, and kg is mass not force) then T = 7500 * 9.81 = 73575N = 73.6kN