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A student investigates moments using a beam placed on a pivot as shown in the diagram - OCR Gateway - GCSE Physics - Question 9 - 2023 - Paper 3

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A student investigates moments using a beam placed on a pivot as shown in the diagram. The student holds the beam. 20 cm 20 cm 40 c... show full transcript

Worked Solution & Example Answer:A student investigates moments using a beam placed on a pivot as shown in the diagram - OCR Gateway - GCSE Physics - Question 9 - 2023 - Paper 3

Step 1

What happens to the beam when the student releases it?

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Answer

To analyze the situation, we need to calculate the net moment acting on the beam around the pivot point.

  1. Moment due to 100 N force: The distance from the pivot is 20 cm.

    Moment=Force×Distance=100N×0.2m=20 N m \text{Moment} = \text{Force} \times \text{Distance} = 100 N \times 0.2 m = 20 \text{ N m}

  2. Moment due to 10 N force: This force is located at the pivot point (0 cm), thus its moment is:

    Moment=10N×0=0 N m\text{Moment} = 10 N \times 0 = 0 \text{ N m}

  3. Moment due to 30 N force: The distance from the pivot is 40 cm.

    Moment=30N×0.4m=12 N m \text{Moment} = 30 N \times 0.4 m = 12 \text{ N m}

Next, we determine the direction of these moments:

  • The moment created by the 100 N force is anti-clockwise.
  • The moment created by the 30 N force is clockwise.

Summing them up:

  • Net moment = Anti-clockwise moment - Clockwise moment

    Net Moment=20 N m12 N m=8 N m\text{Net Moment} = 20 \text{ N m} - 12 \text{ N m} = 8 \text{ N m}

Since the beam rotates anti-clockwise with a net moment greater than zero, the correct answer is D: It stays in equilibrium.

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