Moments Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Moments questions. See exactly how to solve problems on moment, turning-effect, weight, resultant.

momentturning-effectweightresultantequilibriummoments
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A force of 10 N10\,\text{N} acts at a perpendicular distance of 2 m2\,\text{m} from a pivot. Find the magnitude of the moment about the pivot.

Worked solution

  1. Recall the moment formula

    M=FdM=Fd

    The moment of a force about a point equals the force times the perpendicular distance.

  2. Substitute the force and distance

    M=10×2M=10\times 2

    Insert the given values.

  3. State the moment

    M=20 N mM=20\,\text{N m}

    This is the required moment.

Answer
M=20 N mM=20\,\text{N m}
Question 2
2 markseasy
A force of 15 N15\,\text{N} acts at a perpendicular distance of 3 m3\,\text{m} from a pivot. Find the magnitude of the moment about the pivot.

Worked solution

  1. Recall the moment formula

    M=FdM=Fd

    The moment of a force about a point equals the force times the perpendicular distance.

  2. Substitute the force and distance

    M=15×3M=15\times 3

    Insert the given values.

  3. State the moment

    M=45 N mM=45\,\text{N m}

    This is the required moment.

Answer
M=45 N mM=45\,\text{N m}
Question 3
2 markseasy
A force of 8 N8\,\text{N} acts at a perpendicular distance of 4 m4\,\text{m} from a pivot. Find the magnitude of the moment about the pivot.

Worked solution

  1. Recall the moment formula

    M=FdM=Fd

    The moment of a force about a point equals the force times the perpendicular distance.

  2. Substitute the force and distance

    M=8×4M=8\times 4

    Insert the given values.

  3. State the moment

    M=32 N mM=32\,\text{N m}

    This is the required moment.

Answer
M=32 N mM=32\,\text{N m}
Question 4
2 markseasy
A particle of mass 5 kg5\,\text{kg} hangs at a perpendicular distance of 2 m2\,\text{m} from a pivot. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the magnitude of the moment of its weight about the pivot.

Worked solution

  1. Find the weight of the particle

    W=mg=5×9.8W=mg=5\times 9.8

    Weight equals mass times gg.

  2. Recall the moment formula

    M=FdM=Fd

    Use the weight as the force acting at the given distance.

  3. State the moment

    M=98 N mM=98\,\text{N m}

    This is the required moment.

Answer
M=98 N mM=98\,\text{N m}
Question 5
2 markseasy
A particle of mass 4 kg4\,\text{kg} hangs at a perpendicular distance of 3 m3\,\text{m} from a pivot. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the magnitude of the moment of its weight about the pivot. Give your answer to 3 significant figures.

Worked solution

  1. Find the weight of the particle

    W=mg=4×9.8W=mg=4\times 9.8

    Weight equals mass times gg.

  2. Recall the moment formula

    M=FdM=Fd

    Use the weight as the force acting at the given distance.

  3. State the moment

    M=118 N mM=118\,\text{N m}

    This is the required moment.

Answer
M=118 N mM=118\,\text{N m}

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