Question

Seema opens a nut with the help of a spanner of length 20 cm by applying a force of 100 Newton. what should be the minimum length of the spanner if her younger brother Jai opens it with a force of 40 Newton.​

Answer 1

Solution-

Given:

• $\sf{{d_1}=20cm=0.2m}$
• $\sf{{d_2}=100N}$

$\sf{∴ M ={F_1}×{d_1}=100×0.2=20Nm}$

This means that to open the nut, the moment of force should be 20 Nm.

Now for the same turning effect, Jai applies a force $\sf{{F_2}=40 N}$

Therefore, the length of the wrench needed to produce the necessary turning effect will be calculated as, $\sf{M=F_2}×{d_2}.$

$\implies\sf{d_2}=\dfrac{M}{F_2}=\dfrac{20\:Nm}{40\:N}=0.5\:m$

Hence Jai should use a spanner of length $\sf\red{0.5\:m\:or\:50\:cm}$to open the nut.

Answer 2

Solution-

Given:

\sf{{d_1}=20cm=0.2m}d1=20cm=0.2m

\sf{{d_2}=100N}d2=100N

\sf{∴ M ={F_1}×{d_1}=100×0.2=20Nm}∴M=F1×d1=100×0.2=20Nm

This means that to open the nut, the moment of force should be 20 Nm.

Now for the same turning effect, Jai applies a force \sf{{F_2}=40 N}F2=40N

Therefore, the length of the wrench needed to produce the necessary turning effect will be calculated as, \sf{M=F_2}×{d_2}.M=F2×d2.

\implies\sf{d_2}=\dfrac{M}{F_2}=\dfrac{20\:Nm}{40\:N}=0.5\:m⟹d2=F2M=40N20Nm=0.5m

Hence Jai should use a spanner of length \sf\red{0.5\:m\:or\:50\:cm}0.5mor50cm to open the nut.