Question

An effort of 75N is required to lift the load of 300N. Calculate the effort distance if the distance of the load from the fulcrum is 25cm​

Answer 1

Using the principle of moments, the effort distance required to lift a 300N load with a 75N effort when the load is 25cm from the fulcrum is calculated to be 1.0 meter.

Step-by-step explanation:

The question involves calculating the distance an effort force is applied in a simple lever system in physics. When a 75N effort is required to lift a 300N load and the distance of the load from the fulcrum is 25cm, we can use the principle of moments which states that the moment of the effort force about the fulcrum should equal the moment of the load force about the fulcrum for the system to be in equilibrium. The moment is the product of force and distance from the fulcrum (moment = force x distance).

The effort moment (Me) is equal to the load moment (ML), so:

Me = ML

(Effort x Effort Distance) = (Load x Load Distance)

(75N x Effort Distance) = (300N x 25cm)

To find the Effort Distance, we can rearrange the equation:

Effort Distance = (Load x Load Distance) / Effort

Effort Distance = (300N x 25cm) / 75N

Effort Distance = 100cm or 1.0m

Therefore, the effort distance required is 1.0 meter.

Answer 2

`100\; {\rm cm}`, assuming that the force from the load is perpendicular to the lever.

Step-by-step explanation:

The torque of a force on a lever is:

`\tau = F\, r\, \sin(\theta)`,

Where:

• 
  `F` is the magnitude of the force,
• 
  `r` is the distance between the fulcrum and the point where this force is applied, and
• 
  `\theta` is the angle between this force and the lever.

If a lever is in equilibrium, torques on the lever should be balanced.

Let
`F_(1) = 75\; {\rm N}` denote the external force on the lever. Assume that this external force is perpendicular to the lever, such that
`\theta_(1) = 90^(\circ)` is the angle between this force and the lever. Let
`r_(1)` denote the distance between the fulcrum and where this force is applied.

Let
`F_(2) = 300\; {\rm N}` denote the force from the load on the lever. Assuming that this force is also perpendicular to the lever but in the downward direction, the angle between this force and the lever would be
`\theta_(2) = (-90^(\circ))`. Let
`r_(2) = 25\; {\rm cm}` denote the distance between the fulcrum and where this force is applied.

The torque from the external force would be
`\tau_(1) = F_(1)\, r_(1)\, \sin(\theta_(1))`.

The torque from the load would be
`\tau_(2) = F_(2)\, r_(2)\, \sin(\theta_(2))`.

Since torques on this lever should be balanced,
`\tau_(1) + \tau_(2) = 0`. Solve this equation for
`r_(1)`:

`F_(1)\, r_(1)\, \sin(\theta_(1)) + F_(2)\, r_(2)\, \sin(\theta_(2)) = 0`.

`\begin{aligned} r_(1) &= (-F_(2)\, r_(2)\, \sin(\theta_(2)))/(F_(1)\, \sin(\theta_(1))) \\ &= \frac{-(300\; {\rm N})\, (25\; {\rm cm})\, \sin(-90^(\circ))}{(75\; {\rm N})\, \sin(90^(\circ))} \\ &= \frac{(300\; {\rm N})\, (25\; {\rm cm})}{(75\; {\rm N})} \\ &= 100\; {\rm cm}\end{aligned}`.

In other words, if the
`75\; {\rm N}` external force is perpendicular to the lever, that force should be applied at a distance of
`100\; {\rm cm}` from the fulcrum for the lever to be in equilibrium.