Question

A one meter scale AB is balanced horizontally across a knife-edge as shown in the figure

the force on the knife-edge

Answer 1

Step 1: Given data

Length of the scale = 100 cm

Mid-point of the scale = 50 cm

The distance between the knife edge and 50g weight = 30 cm

Weight at point A = 50g

Let the mass of the scale is w.

Step 2: Calculating the weight of the scale

Moment of force $\tau =F\times d$ where F is the force, d is the perpendicular distance of line of action of force from the axis of rotation

The sum of the anticlockwise moment on the left arm = The sum of the clockwise moment on the right arm

$$\begin{gathered} Torque,\tau =F\times d \\ \mathrm{Force}\mathrm{on}\mathrm{the}\mathrm{left}\mathrm{arm}\times \mathrm{distance}\mathrm{from}\mathrm{the}\mathrm{fixed}\mathrm{axis}\mathrm{for}\mathrm{left}\mathrm{arm}=\mathrm{Force}\mathrm{on}\mathrm{the}\mathrm{right}\mathrm{arm}\times \mathrm{distance}\mathrm{from}\mathrm{the}\mathrm{fixed}\mathrm{axis}\mathrm{for}\mathrm{right}\mathrm{arm} \\ \mathrm{w}\times \left(50-30\right)cm=50\mathrm{g}\times \left(100-30\right)cm \\ \mathrm{w}\times 20cm=50g\times 70cm \\ w=\frac{50g\times 70cm}{20cm} \\ \mathrm{w}=\frac{3500}{20} \\ \mathrm{w}=175\mathrm{g} \end{gathered}$$

So, the mass of the scale is 175 g

Step 3: Calculating the force on the knife-edge

The total force on the knife-edge=$F=\left(m_1+m_2\right)g$where m₁ and m₂ are masses, g is acceleration due to gravity

$F=(\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{scale}+\mathrm{weight}\mathrm{at}\mathrm{point}\mathrm{A})\mathrm{g}$ =$\left(175+50\right)10=2250N$

Step 4: Final answer

Thus the force on the knife-edge is 2250N.