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A student investigated the laws of equilibrium for a set of co-planar forces acting on a metre stick - Leaving Cert Physics - Question 1 - 2007

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A student investigated the laws of equilibrium for a set of co-planar forces acting on a metre stick. The weight of the metre stick was 1.2 N and its centre of gravi... show full transcript

Worked Solution & Example Answer:A student investigated the laws of equilibrium for a set of co-planar forces acting on a metre stick - Leaving Cert Physics - Question 1 - 2007

Step 1

How did the student know the metre stick was in equilibrium?

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Answer

The student knew the metre stick was in equilibrium because it was balanced, showing no net movement. This indicates that the resultant force was zero, meaning that the upward forces equaled the downward forces.

Step 2

Copy the diagram and show all the forces acting on the metre stick.

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Answer

In the diagram, the forces should be illustrated as follows:

  • Upward forces: 10 N at the right end and 15 N positioned at 80 cm.
  • Downward force: 1.2 N at the centre of gravity, directly at the 50 cm mark. The forces should be clearly labeled.

Step 3

Find the total upward force acting on the metre stick.

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Answer

The total upward force acting on the metre stick can be calculated by adding the upward forces: extTotalUpwardForce=10extN+15extN=25extN ext{Total Upward Force} = 10 ext{ N} + 15 ext{ N} = 25 ext{ N}

Step 4

Find the total downward force acting on the metre stick.

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Answer

The total downward force acting on the metre stick is simply the weight of the metre stick, which is: extTotalDownwardForce=1.2extN ext{Total Downward Force} = 1.2 ext{ N}. Thus, if there are no other downward forces mentioned in the question, this remains the total.

Step 5

Explain how these values verify one of the laws of equilibrium.

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Answer

In equilibrium, the sum of upward forces must equal the sum of downward forces. Since: extTotalUpwardForce=25extN ext{Total Upward Force} = 25 ext{ N} andextTotalDownwardForce=1.2extN ext{Total Downward Force} = 1.2 ext{ N}, there seems to be an error since these values should be equal for equilibrium. Thus, it denotes that either the diagram has additional forces not accounted for, or there is a mistake in counting the forces.

Step 6

Find the sum of the anticlockwise moments of the upward forces about the O mark.

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Answer

The sum of the anticlockwise moments can be calculated by:

  • Moment due to 15 N at 80 cm: extMoment=15extNimes(80extcm0extcm)=15extNimes80extcm=1200extNcm ext{Moment} = 15 ext{ N} imes (80 ext{ cm} - 0 ext{ cm}) = 15 ext{ N} imes 80 ext{ cm} = 1200 ext{ N cm}
  • Moment due to 10 N at 100 cm:
    extMoment=10extNimes(100extcm0extcm)=10extNimes100extcm=1000extNcmext{Moment} = 10 ext{ N} imes (100 ext{ cm} - 0 ext{ cm}) = 10 ext{ N} imes 100 ext{ cm} = 1000 ext{ N cm}

Thus, extTotalAnticlockwiseMoment=1200extNcm+1000extNcm=2200extNcm ext{Total Anticlockwise Moment} = 1200 ext{ N cm} + 1000 ext{ N cm} = 2200 ext{ N cm}

Step 7

Find the sum of the clockwise moments of the downward forces about the O mark.

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Answer

The downward forces create clockwise moments. The only downward force in this scenario is the 1.2 N acting at the 50 cm mark: extClockwiseMoment=1.2extNimes(50extcm0extcm)=1.2extNimes50extcm=60extNcm ext{Clockwise Moment} = 1.2 ext{ N} imes (50 ext{ cm} - 0 ext{ cm}) = 1.2 ext{ N} imes 50 ext{ cm} = 60 ext{ N cm}

Step 8

Explain how these values verify the other law of equilibrium.

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Answer

In equilibrium, the total anticlockwise moments must equal the total clockwise moments. We have:

  • Anticlockwise Moment: 2200 N cm
  • Clockwise Moment: 60 N cm

These values do not equal, indicating that the system is not in true equilibrium. The principle states that: extTotalAnticlockwiseMoment=extTotalClockwiseMoment ext{Total Anticlockwise Moment} = ext{Total Clockwise Moment} When these values are not equal, it highlights an error in force applications or placements.

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