Question

A UNIFORM METRE RULE AB PIVOTED AT ITS A AT ZERO MARK AND SUPPORTED AT THE OTHER END B BY A SPRING BALANCE WHEN A WEIGHT OF 40kgf IS SUSPENDED AT ITS 40cm MARK.THIS RULE STAYS HORIZONTAL.FIND THE READING OF THE SPRING BALANCE WHEN THE RULE IS OF
1) NEGLIGIBLE MASS
2)MASS 20kg

Answer 1

Given that,

Weight = 40 kgf

Length = 40 cm

(I). We need to calculate the force

Using principle of moment

Anticlockwise moment=clockwise moment

$F\times 1=40\times0.4$

$F=16\ kgf$

So, the reading of the spring balance is 16 kgf,

We know that,

1 kgf = 9.8 N

$16\ kgf = 16\times9.8 = 156.8\ N$

(II), If mass of 20 kg

Then, weight will be

$W=20\times9.8\ N$

$W=196\ N$

We need to calculate the force

Using principle of moment

Anticlockwise moment=clockwise moment

$F\times 1=(40\times9.8+196)\times 0.4$

$F=235.2\ N$

Hence, (I). The reading of the spring balance is 156.8 N.

(II). The reading of the spring balance is 235.2 N.