Statics of rigid bodies Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Statics of rigid bodies questions. See exactly how to solve problems on non-uniform-rod, two-strings, moments, beam.

non-uniform-rodtwo-stringsmomentsbeamsupportsladder
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
A non-uniform rod ABAB of weight 4040 N and length 44 m rests horizontally in equilibrium, suspended by two vertical light strings attached at the ends AA and BB. The centre of mass of the rod is a distance 1.51.5 m from AA. Find the tension in the string at AA.

Worked solution

  1. Take moments about AA

    TB×4=40×1.5T_B\times 4=40\times 1.5

    Taking moments about AA removes TAT_A from the equation.

  2. Solve for the tension at BB

    TB=15 NT_B=15\ \text{N}

    Divide the total moment about AA by the length of the rod.

  3. State the answer

    TA=25T_A=25

    This is the required tension, in newtons.

Answer
TA=25 NT_A=25\ \text{N}
Question 2
2 markseasy
A non-uniform rod ABAB of weight 6060 N and length 55 m rests horizontally in equilibrium, suspended by two vertical light strings attached at the ends AA and BB. The centre of mass of the rod is a distance 22 m from AA. Find the tension in the string at BB.

Worked solution

  1. Take moments about AA

    TB×5=60×2T_B\times 5=60\times 2

    Taking moments about AA removes TAT_A from the equation.

  2. Solve for the tension at BB

    TB=24 NT_B=24\ \text{N}

    Divide the total moment about AA by the length of the rod.

  3. Substitute back to find the tension at AA

    TA=60−24=36 NT_A=60-24=36\ \text{N}

    The remaining tension follows from the vertical equation.

  4. State the answer

    TB=24T_B=24

    This is the required tension, in newtons.

Answer
TB=24 NT_B=24\ \text{N}
Question 3
2 markseasy
A non-uniform rod ABAB of weight 5050 N and length 44 m hangs horizontally in equilibrium from two vertical light strings attached at AA and BB. The tension in the string at BB is 3030 N. Find the distance of the centre of mass of the rod from AA.

Worked solution

  1. Find the tension at AA

    TA=50−30=20 NT_A=50-30=20\ \text{N}

    Subtract the known tension from the weight.

  2. Take moments about AA

    30×4=50×xˉ30\times 4=50\times \bar{x}

    The moment of the weight about AA equals that of TBT_B.

  3. Solve for the distance of the centre of mass

    xˉ=30×450=2.4 m\bar{x}=\frac{30\times 4}{50}=2.4\ \text{m}

    Divide the moment of TBT_B about AA by the weight.

  4. State the distance

    xˉ=2.4 m\bar{x}=2.4\ \text{m}

    This locates the centre of mass along the rod from AA.

Answer
xˉ=2.4 m\bar{x}=2.4\ \text{m}
Question 4
2 markseasy
A uniform beam ABAB of weight 200200 N and length 66 m rests horizontally on two supports, one at CC a distance 11 m from AA and the other at DD a distance 55 m from AA. Find the magnitude of the reaction at the support CC.

Worked solution

  1. Take moments about CC

    RD (5−1)=200 (12×6−1)R_D\,(5-1)=200\,(\tfrac{1}{2}\times 6-1)

    Moments about CC eliminate RCR_C from the equation.

  2. Solve for the reaction at DD

    RD=100 NR_D=100\ \text{N}

    Divide the total moment about CC by the distance between the supports.

  3. State the answer

    RC=100 NR_C=100\ \text{N}

    This is the required support reaction.

Answer
RC=100 NR_C=100\ \text{N}
Question 5
2 markseasy
A uniform beam ABAB of weight 120120 N and length 88 m rests horizontally on two supports, one at CC a distance 22 m from AA and the other at DD a distance 66 m from AA. Find the magnitude of the reaction at the support DD.

Worked solution

  1. Take moments about CC

    RD (6−2)=120 (12×8−2)R_D\,(6-2)=120\,(\tfrac{1}{2}\times 8-2)

    Moments about CC eliminate RCR_C from the equation.

  2. Resolve vertically for the beam

    RC+RD=120 NR_C+R_D=120\ \text{N}

    The two support reactions carry the whole load.

  3. State the answer

    RD=60 NR_D=60\ \text{N}

    This is the required support reaction.

Answer
RD=60 NR_D=60\ \text{N}

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