A block of mass 20 kg is suspended through two spring balances with negligible mass as shown in figure. What will be the readings in the upper and lower balance respectively?
Show Hint
When force measuring instruments with negligible mass are connected in a series line, the same force propagates through all of them . Think of it like water pressure in a pipe; at any point without leaks/addition, the pressure is continuous. A common mistake is to add the masses or divide the force. Only one weight creates tension, which is then measured multiple times.
Concept:
A spring balance measures the tension acting through it. When two spring balances are connected in series, the tension throughout the string/balance assembly remains constant if the balances themselves are massless.
Step 1: {Analyze the tension in the lower spring balance.}
The lower balance is directly supporting the 20 kg mass. The tension $T$ in the lower balance is:
$$T = mg = 20 \times g$$
The reading is shown as the mass equivalent, which is 20 kg.
Step 2: {Analyze the tension in the upper spring balance.}
Since the balances have negligible mass, the upper balance supports the lower balance plus the mass.
$$\text{Total mass below upper balance} = m_{\text{lower balance}} + m_{\text{block}}$$
$$\text{Total mass} = 0 \text{ kg} + 20 \text{ kg} = 20 \text{ kg}$$
Step 3: {Apply the principle of tension in series.}
In a series connection of massless components, the force (tension) is transmitted undiminished. Thus, the tension $T$ is the same for both balances.
$$\text{Reading}_1 = \text{Reading}_2 = 20 \text{ kg}$$
