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In an experiment to verify the laws of equilibrium, the centre of gravity and the weight of a metre stick were found - Leaving Cert Physics - Question 1 - 2016

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In an experiment to verify the laws of equilibrium, the centre of gravity and the weight of a metre stick were found. The centre of gravity of the stick was at the 5... show full transcript

Worked Solution & Example Answer:In an experiment to verify the laws of equilibrium, the centre of gravity and the weight of a metre stick were found - Leaving Cert Physics - Question 1 - 2016

Step 1

how (i) the centre of gravity was found

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Answer

The centre of gravity was determined by balancing the metre stick horizontally on a pivot. This was achieved by suspending the stick from a thread at the point where it remained stable, indicating that the forces acting on it were balanced.

Step 2

how (ii) the weight of the metre stick was found

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Answer

The weight of the metre stick was measured using a Newton balance or weighing scales. The mass of the stick was obtained, and then this mass was multiplied by the acceleration due to gravity (g = 9.81 m/s²) to calculate the weight.

Step 3

how (iii) the upward forces and downward forces were determined

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Answer

The upward forces acting on the metre stick were determined by using Newton balances to measure the forces applied upward, while the downward forces included the known weights attached to the stick, as well as its weight, which was previously calculated.

Step 4

Give one possible reason why the centre of gravity is not at the 50.0 cm mark

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Answer

One possible reason the centre of gravity is not at the 50.0 cm mark is that the stick may have been worn on one side, or there could be a hole in one side, making it not uniform in its mass distribution.

Step 5

Use the data given to calculate (g) the net force acting on the metre stick

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Answer

The net force acting on the metre stick can be calculated by summing the upward forces and the downward forces.

Upward forces: extTotalUpwardForce=3.9extN+4.1extN=8.0extN ext{Total Upward Force} = 3.9 ext{ N} + 4.1 ext{ N} = 8.0 ext{ N}

Downward forces: extTotalDownwardForce=2extN+3extN+2extN+1.1extN=8.1extN ext{Total Downward Force} = 2 ext{ N} + 3 ext{ N} + 2 ext{ N} + 1.1 ext{ N} = 8.1 ext{ N}

Net vertical force can then be calculated as: extNetVerticalForce=extUpwardForceextDownwardForce=8.0extN8.1extN=0.1extN ext{Net Vertical Force} = ext{Upward Force} - ext{Downward Force} = 8.0 ext{ N} - 8.1 ext{ N} = -0.1 ext{ N}

Step 6

Use the data given to calculate (ii) the sum of the moments about the 40 cm mark of the metre stick

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Answer

The moment about a point is calculated using the formula: extMoment=extForceimesextDisplacement ext{Moment} = ext{Force} imes ext{Displacement}

Clockwise moments at the 40 cm mark: extClockwiseMoment=(2extNimes0.52)+(1.1extNimes0.102)+(3.9extNimes0.004)=1.3082extNm ext{Clockwise Moment} = (2 ext{ N} imes 0.52) + (1.1 ext{ N} imes 0.102) + (3.9 ext{ N} imes 0.004) = 1.3082 ext{ Nm}

Anticlockwise moments: extAnticlockwiseMoment=(2extNimes0.24)+(4.1extNimes0.20)=1.3extNm ext{Anticlockwise Moment} = (2 ext{ N} imes 0.24) + (4.1 ext{ N} imes 0.20) = 1.3 ext{ Nm}

Summing both moments: extSumofMoments=0.01extNm ext{Sum of Moments} = 0.01 ext{ Nm}

Step 7

Explain how your calculations verify the laws of equilibrium

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Answer

According to the laws of equilibrium, the net vertical force should equal zero, and the sum of the moments about any point should also equal zero. In this case, the net vertical force was calculated to be -0.1 N, which indicates an imbalance, suggesting that the system is not in equilibrium. Additionally, the sum of the moments about the point was found to be close to zero, which verifies the concept of balancing moments in equilibrium.

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