In a hydraulic lift force F is applied to balance 10 N load, diameter of effort arm is 14 cm and load arm is 1.4 cm. The F is equal to

Question

A cubic block of mass $ m $ is sliding down on an inclined plane at $ 60^\circ $ with an acceleration of $ \frac{g}{2} $, the value of coefficient of kinetic friction is:

Answer 1

To solve this problem, we need to understand how a hydraulic lift works. A hydraulic lift multiplies the force applied using Pascal's principle, which states that pressure applied to a confined fluid is transmitted undiminished in all directions throughout the fluid, regardless of the area over which it was applied.

Given in the problem:

Force required to balance the load, $ F_{\text{load}} = 10 \, \text{N} $
Diameter of the effort arm, $ d_{\text{effort}} = 14 \, \text{cm} $
Diameter of the load arm, $ d_{\text{load}} = 1.4 \, \text{cm} $

The force applied at the effort arm is balanced by the load at the load arm through the hydraulic lift, which can be expressed by the formula for hydraulic systems:

F_{\text{effort}} \cdot A_{\text{effort}} = F_{\text{load}} \cdot A_{\text{load}}

where:

$ A_{\text{effort}} $ is the area of the effort piston
$ A_{\text{load}} $ is the area of the load piston

Assuming both pistons have circular cross-sections, we can use the formula for the area of a circle $ A = \frac{\pi d^2}{4} $. Therefore,

A_{\text{effort}} = \frac{\pi (d_{\text{effort}})^2}{4}

A_{\text{load}} = \frac{\pi (d_{\text{load}})^2}{4}

Substitute $ A_{\text{effort}} $ and $ A_{\text{load}} $ into the original formula, we get:

F_{\text{effort}} \cdot \frac{\pi (d_{\text{effort}})^2}{4} = 10 \cdot \frac{\pi (d_{\text{load}})^2}{4}

By cancelling the common terms and rearranging for $ F_{\text{effort}} $, we have:

F_{\text{effort}} = 10 \cdot \left(\frac{d_{\text{load}}}{d_{\text{effort}}}\right)^2

Substitute the given values:

F_{\text{effort}} = 10 \cdot \left(\frac{1.4}{14}\right)^2

Calculate the above expression:

F_{\text{effort}} = 10 \cdot \left(\frac{1}{10}\right)^2

F_{\text{effort}} = 10 \cdot \frac{1}{100}

F_{\text{effort}} = 1000 \, \text{N}

Thus, the force $ F $ needed to balance the hydraulic system is 1000 N, which is the correct answer.

Answer 2