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The laws of equilibrium for a set of co-planar forces acting on a metre stick were investigated by a student - Leaving Cert Physics - Question 1 - 2013
Question 1
The laws of equilibrium for a set of co-planar forces acting on a metre stick were investigated by a student. She first found the centre of gravity of the metre stic... show full transcript
Worked Solution & Example Answer:The laws of equilibrium for a set of co-planar forces acting on a metre stick were investigated by a student - Leaving Cert Physics - Question 1 - 2013
Step 1
How did the student find the centre of gravity of the metre stick?
Answer
The student found the centre of gravity by balancing the metre stick horizontally at its centre of gravity point. This was achieved by suspending the stick from a string at a point where the forces acting on it were equal, indicating that the system was stable and balanced.
Step 2
How did the student ensure that the system was at equilibrium?
Answer
The student ensured the system was at equilibrium by adjusting the weights until the net moment around the pivot point was zero. This involved checking that the sum of clockwise moments equaled the sum of anti-clockwise moments.
Step 3
Draw a diagram of the experimental arrangement that the student used.
Answer
The diagram should depict the metre stick suspended horizontally with weights attached at specified positions and supported by two spring balances. This visual representation would include labels to indicate forces and directions.
Step 4
(i) the total of the clockwise moments
Answer
To calculate the total clockwise moments, we use the positions and forces given:
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For 30.3 cm:
Torque = Force × Distance = 5.7 N × (65.4 cm - 50 cm) = 5.7 N × 15.4 cm = 0.878 Nm
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For 80.0 cm:
Torque = 4.0 N × (80.0 cm - 50 cm) = 4.0 N × 30.0 cm = 1.200 Nm
Total clockwise moments = 0.878 + 1.200 = 2.078 Nm.
Step 5
(ii) the total of the anti-clockwise moments
Answer
To calculate the total anti-clockwise moments, we also use the positions and forces given:
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For 11.4 cm:
Torque = 2.0 N × (50 cm - 11.4 cm) = 2.0 N × 38.6 cm = 0.772 Nm
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For 21.8 cm:
Torque = 3.0 N × (50 cm - 21.8 cm) = 3.0 N × 28.2 cm = 0.846 Nm
Total anti-clockwise moments = 0.772 + 0.846 = 1.618 Nm.
Step 6
Explain how these results verify the laws of equilibrium.
Answer
The laws of equilibrium state that for an object to be in equilibrium, the total clockwise moments must equal the total anti-clockwise moments. Given that the calculated total clockwise moments (2.078 Nm) approximates the anti-clockwise moments (1.618 Nm) with some experimental error, this supports the fundamental principle of equilibrium where forces act in balance.