Question

A half-meter rod is pivoted at the center with two weights of 20 gf and 12 gf suspended at a perpendicular distance of 6 cm and 10 cm from the pivot respectively as shown alongside.

Is the rod in equilibrium?

Answer 1

Step 1:Given

According to the diagram, the pivot is balancing the forces.

Step 2: Formula & solution

Moment of force $\tau =F\times d$ where F is force, d is the distance

Weight on the left arm W = 20gf

Weight on the right arm w = 12gf

Distance between the pivot and 12gf = 10 cm

The force on the left arm:

$$\begin{gathered} \mathrm{weight}\mathrm{on}\mathrm{the}\mathrm{left}\mathrm{arm}\mathrm{W}\times \mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{pivot}\mathrm{and}20\mathrm{gf} \\ =20\mathrm{gf}\times 6\mathrm{cm} \\ =120\mathrm{gf}-\mathrm{cm} \end{gathered}$$

The force on the right arm:

$$\begin{gathered} \mathrm{weight}\mathrm{on}\mathrm{the}\mathrm{right}\mathrm{arm}\mathrm{w}\times \mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{pivot}\mathrm{and}12\mathrm{gf} \\ =12\mathrm{gf}\times 10\mathrm{cm} \\ =120\mathrm{gf}-\mathrm{cm} \end{gathered}$$

The force on the left arm = The force on the right arm

So, the net toque is zero. Therefore, the rod is in equilibrium