Question

Hello!!

A uniform meter scale is in equilibrium as shown in the diagram :

1) Calculate the weight of the meter scale.

2) Which of the following options is correct to keep the ruler in equilibrium when 40 gf wt is shifted to 0cm Mark?

F is shifted towards 0cm.
or
F is shifted towards 100cm. ​

Answer 1

Heya!!

Here's your Answer!!

1) Let the weight of the meter scale be x gf and it acts at the center of gravity ( i.e., 50 cm mark).

Anticlockwise moment = 40 × 25 gf cm

Clockwise moment =x ×20 gf cm

When the meter scale is balanced,

Clockwise moment = Anticlockwise moment

= x × 20 = 40 × 25

= x = 40 × 25/ 20 gf

= = 50 gf

Therefore, Weight of meter scale is 50 gf.

2) F is shifted towards 0 cm.

Glad help you,

it hepls you,

thanks.

Answer 2

Given that,

Weight = 40 gf

According to figure,

(1). We need to calculate the weight of the meter scale

Using formula balance equation

Anti clockwise moment = Clockwise moment

$W\times x=W'\times x'$

Put the value into the formula

$40\times25=W\times20$

$W'=\dfrac{40\times25}{20}$

$W'=50 gf$

(2). We need to calculate the ruler in equilibrium

If F shifted to 0 cm then the torque will be zero.

$\tau=0$

So, F can be shifted to 100 cm.

Hence, (1). The weight of the meter scale is 50 gf.

(2). F is shifted toward 100 cm.