Hard A-Level Moments Questions

Challenging, exam-style A-Level Moments questions with worked solutions. Stretch yourself on the hardest moment, turning-effect, weight, resultant problems.

momentturning-effectweightresultantequilibriummoments
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
Two particles of masses m1m_1 and m2m_2 lie on a straight line at positions x1x_1 and x2x_2. Which expression gives the xx-coordinate of their centre of mass?
Show worked solution

Worked solution

  1. Recall the centre of mass formula

    xˉ=m1x1+m2x2m1+m2\bar{x}=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}

    The centre of mass is a weighted average of positions.

  2. Note the weighting

    heavier mass has more influence\text{heavier mass has more influence}

    Larger masses pull the centre of mass closer.

  3. Recall the modelling assumptions

    rigid body, coplanar forces\text{rigid body, coplanar forces}

    Standard moment problems treat the body as rigid with forces in one plane.

  4. Note the pivot point

    moments are taken about the stated pivot\text{moments are taken about the stated pivot}

    The pivot is the reference point for all distances.

  5. Distinguish force and moment

    force in N, moment in N m\text{force in N, moment in N m}

    A moment is not the same as a force.

  6. Check the units

    N m = N×m\text{N m = N}\times\text{m}

    Moment units are newton metres.

  7. State the equilibrium conditions

    ΣF=0, ΣM=0\Sigma F=0,\ \Sigma M=0

    Complete equilibrium needs both conditions.

  8. Recall the sign convention

    anticlockwise positive\text{anticlockwise positive}

    A consistent sign convention is needed to add moments.

  9. Identify the lever arm

    d is perpendicular to the forced\ \text{is perpendicular to the force}

    Only the perpendicular distance counts.

  10. Summarise the key idea

    turning effect =Fd\text{turning effect }=Fd

    The moment measures the turning effect of a force.

  11. Apply to a uniform rod

    W acts at the centreW\ \text{acts at the centre}

    A uniform rod's weight acts at its midpoint.

  12. Relate to centre of mass

    xˉ=∑mixi∑mi\bar{x}=\dfrac{\sum m_ix_i}{\sum m_i}

    The centre of mass is the weighted average of positions.

  13. Consider tilting

    R=0 at one supportR=0\ \text{at one support}

    Tilting begins when a support reaction becomes zero.

  14. Verify the logic

    check against the definition\text{check against the definition}

    The correct answer follows from the definition.

  15. Select the correct formula

    xˉ=m1x1+m2x2m1+m2\bar{x}=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}

    This is the formula for two particles on a line.

Answer
xˉ=m1x1+m2x2m1+m2\bar{x}=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}
Question 2
8 markschallenging
A uniform beam rests horizontally on two supports and is in equilibrium. Which pair of conditions must hold?
Show worked solution

Worked solution

  1. Recall beam equilibrium

    two conditions\text{two conditions}

    A beam in equilibrium must satisfy both force and moment conditions.

  2. List the conditions

    ΣF=0, ΣM=0\Sigma F=0,\ \Sigma M=0

    Zero resultant force and zero resultant moment.

  3. Recall the modelling assumptions

    rigid body, coplanar forces\text{rigid body, coplanar forces}

    Standard moment problems treat the body as rigid with forces in one plane.

  4. Note the pivot point

    moments are taken about the stated pivot\text{moments are taken about the stated pivot}

    The pivot is the reference point for all distances.

  5. Distinguish force and moment

    force in N, moment in N m\text{force in N, moment in N m}

    A moment is not the same as a force.

  6. Check the units

    N m = N×m\text{N m = N}\times\text{m}

    Moment units are newton metres.

  7. State the equilibrium conditions

    ΣF=0, ΣM=0\Sigma F=0,\ \Sigma M=0

    Complete equilibrium needs both conditions.

  8. Recall the sign convention

    anticlockwise positive\text{anticlockwise positive}

    A consistent sign convention is needed to add moments.

  9. Identify the lever arm

    d is perpendicular to the forced\ \text{is perpendicular to the force}

    Only the perpendicular distance counts.

  10. Summarise the key idea

    turning effect =Fd\text{turning effect }=Fd

    The moment measures the turning effect of a force.

  11. Apply to a uniform rod

    W acts at the centreW\ \text{acts at the centre}

    A uniform rod's weight acts at its midpoint.

  12. Relate to centre of mass

    xˉ=∑mixi∑mi\bar{x}=\dfrac{\sum m_ix_i}{\sum m_i}

    The centre of mass is the weighted average of positions.

  13. Consider tilting

    R=0 at one supportR=0\ \text{at one support}

    Tilting begins when a support reaction becomes zero.

  14. Verify the logic

    check against the definition\text{check against the definition}

    The correct answer follows from the definition.

  15. Select the correct pair

    ΣF=0, ΣM=0\Sigma F=0,\ \Sigma M=0

    Both conditions are required.

Answer
ΣF=0, ΣM=0\Sigma F=0,\ \Sigma M=0
Question 3
8 markschallenging
A plank rests on a table with part of it overhanging. Which condition describes the plank being on the point of tilting about the edge of the table?
Show worked solution

Worked solution

  1. Recall the tilting condition

    R=0 at one supportR=0\ \text{at one support}

    Tilting begins when one support reaction becomes zero.

  2. Relate to the centre of mass

    weight line passes through the remaining support\text{weight line passes through the remaining support}

    The body is on the point of tilting when the weight acts through the pivot.

  3. Recall the modelling assumptions

    rigid body, coplanar forces\text{rigid body, coplanar forces}

    Standard moment problems treat the body as rigid with forces in one plane.

  4. Note the pivot point

    moments are taken about the stated pivot\text{moments are taken about the stated pivot}

    The pivot is the reference point for all distances.

  5. Distinguish force and moment

    force in N, moment in N m\text{force in N, moment in N m}

    A moment is not the same as a force.

  6. Check the units

    N m = N×m\text{N m = N}\times\text{m}

    Moment units are newton metres.

  7. State the equilibrium conditions

    ΣF=0, ΣM=0\Sigma F=0,\ \Sigma M=0

    Complete equilibrium needs both conditions.

  8. Recall the sign convention

    anticlockwise positive\text{anticlockwise positive}

    A consistent sign convention is needed to add moments.

  9. Identify the lever arm

    d is perpendicular to the forced\ \text{is perpendicular to the force}

    Only the perpendicular distance counts.

  10. Summarise the key idea

    turning effect =Fd\text{turning effect }=Fd

    The moment measures the turning effect of a force.

  11. Apply to a uniform rod

    W acts at the centreW\ \text{acts at the centre}

    A uniform rod's weight acts at its midpoint.

  12. Relate to centre of mass

    xˉ=∑mixi∑mi\bar{x}=\dfrac{\sum m_ix_i}{\sum m_i}

    The centre of mass is the weighted average of positions.

  13. Consider tilting

    R=0 at one supportR=0\ \text{at one support}

    Tilting begins when a support reaction becomes zero.

  14. Verify the logic

    check against the definition\text{check against the definition}

    The correct answer follows from the definition.

  15. Select the correct statement

    reaction at one support is zero\text{reaction at one support is zero}

    This is the standard tilting criterion.

Answer
reaction at the table edge support is zero\text{reaction at the table edge support is zero}
Question 4
8 markschallenging
A uniform horizontal beam of length 12 m12\,\text{m} and weight 60 N60\,\text{N} rests on smooth supports at 1 m1\,\text{m} and 11 m11\,\text{m} from its left end. Additional loads of 30 N30\,\text{N} at 5 m5\,\text{m} and 20 N20\,\text{N} at 9 m9\,\text{m} act on the beam. Find the vertical reaction at support BB.
Show worked solution

Worked solution

  1. Model the uniform beam with weight at its centre

    W=60 N at L2=6 mW=60\,\text{N}\ \text{at }\dfrac{L}{2}=6\,\text{m}

    A uniform beam has its weight at its midpoint.

  2. Write the equilibrium equations

    RA+RB=W+ΣF, RB(dB−dA)=moments about AR_A+R_B=W+\Sigma F,\ R_B(d_B-d_A)=\text{moments about }A

    Resolve vertically and take moments about support AA.

  3. State the support positions

    dA=1 m, dB=11 md_A=1\,\text{m},\ d_B=11\,\text{m}

    The supports are at these distances from the left end.

  4. List the additional loads

    30 Nat5 m, 20 Nat9 m30\,\text{N} at 5\,\text{m},\ 20\,\text{N} at 9\,\text{m}

    These are the other vertical forces on the beam.

  5. Calculate moments about support AA

    RB(dB−dA)=580R_B(d_B-d_A)=580

    The moment from RBR_B balances all other moments about AA.

  6. Solve for RBR_B

    RB=58 NR_B=58\,\text{N}

    Divide the total moment by the span between supports.

  7. Use vertical resolution to find RAR_A

    RA=(W+ΣF)−RBR_A=(W+\Sigma F)-R_B

    The two reactions sum to the total downward force.

  8. Evaluate RAR_A

    RA=52 NR_A=52\,\text{N}

    Subtract RBR_B from the total load.

  9. Check the moment balance about BB

    RA(dB−dA)=moments about BR_A(d_B-d_A)=\text{moments about }B

    An alternative moment equation should give the same values.

  10. Verify RA+RBR_A+R_B equals the total load

    52+58=11052+58=110

    The reactions sum to the total downward force.

  11. Note the beam length

    L=12 mL=12\,\text{m}

    The beam extends from 00 to LL.

  12. Recall the equilibrium conditions

    ΣF=0, ΣM=0\Sigma F=0,\ \Sigma M=0

    A body in equilibrium has zero resultant force and moment.

  13. State the total downward load

    W+ΣF=110 NW+\Sigma F=110\,\text{N}

    This is the combined weight and any extra loads.

  14. Summarise the reactions

    RA=52 N, RB=58 NR_A=52\,\text{N},\ R_B=58\,\text{N}

    These are the support reactions.

  15. State the reaction at BB

    RB=58 NR_B=58\,\text{N}

    This is the reaction at support BB.

Answer
RB=58 NR_B=58\,\text{N}
Question 5
8 markschallenging
A uniform plank of length 5 m5\,\text{m} and weight 15 N15\,\text{N} rests on a horizontal table with 1.5 m1.5\,\text{m} overhanging the edge. A particle of weight 35 N35\,\text{N} is placed on the plank. Find the distance of the particle from the left end when the plank is on the point of tilting about the edge. Give your answer to 3 significant figures.
Show worked solution

Worked solution

  1. State the tilting condition

    Rleft=0R_{\text{left}}=0

    The plank is on the point of tilting when the reaction at the left support becomes zero.

  2. Relate tilting to the centre of mass

    vertical through COM passes through the pivot\text{vertical through COM passes through the pivot}

    At the point of tilting, the weight acts through the pivot.

  3. Find the total weight

    Wtot=35+15=50 NW_{\text{tot}}=35+15=50\,\text{N}

    Add the particle weight and the plank weight.

  4. Write the centre of mass position

    xˉ=Wpd+Wb⋅L2Wtot\bar{x}=\dfrac{W_pd+W_b\cdot\frac{L}{2}}{W_{\text{tot}}}

    The combined centre of mass is the weighted average of the positions.

  5. Set the centre of mass at the pivot

    xˉ=Lp\bar{x}=L_p

    At the point of tilting, the COM is directly above the pivot.

  6. Substitute the pivot position

    Lp=1.5 mL_p=1.5\,\text{m}

    Insert the distance of the pivot from the left end.

  7. Rearrange for the particle position

    d=Lp+WbWp(Lp−L2)d=L_p+\dfrac{W_b}{W_p}\left(L_p-\dfrac{L}{2}\right)

    Make dd the subject.

  8. Substitute the values

    d=1.07d=1.07

    Insert the numerical values.

  9. Evaluate the critical position

    d=1.07 md=1.07\,\text{m}

    This is the position of the particle when the plank is about to tilt.

  10. Interpret the result

    further right would cause tilting\text{further right would cause tilting}

    If the particle moves beyond this point, the plank tilts.

  11. Note the plank length

    L=5 mL=5\,\text{m}

    The plank extends from 00 to LL.

  12. Recall the uniform plank assumption

    Wb acts at L2W_b\ \text{acts at }\dfrac{L}{2}

    A uniform plank has its weight at its centre.

  13. Check the centre of mass

    xˉ=1.5 m\bar{x}=1.5\,\text{m}

    At this position, the COM is at the pivot.

  14. Summarise

    d=1.07 md=1.07\,\text{m}

    The particle must be at this distance for the plank to be on the point of tilting.

  15. State the particle position

    d=1.07 md=1.07\,\text{m}

    This is the critical position for tilting.

Answer
d=1.07 md=1.07\,\text{m}

Unlock 29 more Moments questions

Create a free account to work through every A-Level Moments question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Moments practice

Practice QuestionsWorked Solutions

Related Mechanics topics

Applications of forcesConnected particlesConstant accelerationForces and frictionForces and Newton’s lawsFurther kinematics