Question

3. A machine can open a nut by applying 150 N force while using a lever handle of 40 cm length. How long handle is required if he wants to open it by applying a force of only 50 N?​

Answer 1

Answer:

1m

Explanation :

Given :

Force applied by mechanic, => 150N

Length of lever, => 40cm

To Find :

Length $\sf{}(l_2)$ required to apply 50N,

Solution :

Equate the torque in both cases:-

$\sf{}f_1l_1=f_2l_2$

$\sf{}150\times 0.4 =l_2\times 60$

$\sf{}60 =l_2\times 60$

$\sf{}l_2=\dfrac{60}{60}$

$\sf{}l_2=1m$

Hence,

1m length is required to apply a force of 50N

☆Additional Information☆

• Torque is a force that can cause rotation in an object.

• The point where the object rotates is known as the axis of rotation. Mathematically, torque can be written as T = F * r * sin(theta), and it has units of Newton-meters.

• It is mainly denoted with a capital “T,” but the correct symbol is the Greek letter tau, “τ.”

• Torque has the dimension of force times distance, symbolically $\sf{} L^2MT^{-2}$

Answer 2

Answer:

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Step-by-step explanation: