Question

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

Answer 1

Answer:–

• 1.2 m

Explaination:–

This question should be divided into two parts. Let us divide them and take as first and second part.

First part:

What is given?

• Force applied = 150N
• Distance = 40cm

What to do now if know the given things?

• We have to calculate the moment of force.

Formula needed to be applied?

• Moment of force = Force × Perpendicular distance.

Let's solve it now!!

Here we needed to apply the given formula of moment of force.

➟ Moment of force= Force× Perpendicular distance

Now substituting the values.

➟ Moment of force = 150N × 40 cm

Here we needed to convert the distance which is 40 cm into metres.

We know that : 1 metre = 100cm

Therefore, 40/100 = 0.4m

Now again solving.

➟ 150N × 0.4m = 60Nm

Second part:

What is given?

• Force applied = 50 N

Needed to find?

• Length

Let's solve now!!

Let us assume that he used the handle of length be Lm.

Therefore, again calculating the moment of force in order to calculate the length.

➟ Moment of force = 50 N × Lm (the assumed value)

➟ 50 N × Lm = 50L Nm

Here we would divide the moment of force which we calculated in the first part from the moment of force calculated in the second part.

➟ Length = 60 Nm / 50 L Nm

➟ Length = 60/50

➟ Length = 6/5

➟ Length = 1.2

Thus,

• Length of handle required is 1.2 m.

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Hope it helps :)