Question

The effort arm of an ideal lever of first class lever is 40cm long and load arm is
1.2m long. How much effort is required to lift a load of 80N?

Answer 1

Answer:

Load/Effort=Load Arm/Effort Arm.

Therefore, 80/Effort= 120/40.

80×40/120= Effort

2.6N= Effort

E=2.6N

Answer 2

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According to law of levers, the product of load arm and load is equal to product of effort arm and effort. This is denoted by :-

$\bigstar \: load \: arm \times load = effort \: arm \: \times effort$

Now in this case we have been given effort arm of a ideal first class lever and also the load arm. A load of 80N is there on lever, effort required to lift load is not known we have to calculate it.

We have been given that :-

• Effort arm = 40cm = 0.4m
• Load arm = 1.2m
• Load = 80N
• Effort = ?

Now if we input these values in law of levers expression we get :-

$\mapsto \: 1.2 \times 80 = 0.4e$

$\mapsto \: 96 = 0.4e$

$\mapsto \: e = \frac{96}{0.4}$

$\mapsto \: \huge\cancel \frac{96}{0.4}$

$e = 240n$

Therefore we get effort required to lift the load is 240N

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BY SCIVIBHANSHU

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