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A pole of length 1.7m and mass 15kg is supported by two people. A boy of mass 65kg sits 0.75m from one end and a girl of mass 49kg sits 0.5m from the other end. Calculate the forces exerted by the sup

Physics

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8. a) i.
The principle of moments states that for a body to be in rotational equilibrium, the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point.

8. a) ii.

  1. Identify the forces and distances.

    Let RMR_M be the reaction force from the mother and RFR_F be the reaction force from the father.
    Pole length L=1.7L = 1.7 m.
    Weight of pole WP=150W_P = 150 N, acting at the center of the pole, 0.85 m from each end.
    Weight of boy WB=650W_B = 650 N, sitting 0.75 m from one end (let's assume the mother's end).
    Weight of girl WG=490W_G = 490 N, sitting 0.5 m from the other end (father's end). This means the girl is at 1.7−0.5=1.21.7 - 0.5 = 1.2 m from the mother's end.

  2. Apply the principle of moments about the mother's support (RMR_M).

    Clockwise moments:
    Moment due to boy = WB×0.75=650 N×0.75 m=487.5 NmW_B \times 0.75 = 650 \text{ N} \times 0.75 \text{ m} = 487.5 \text{ Nm}
    Moment due to pole = WP×0.85=150 N×0.85 m=127.5 NmW_P \times 0.85 = 150 \text{ N} \times 0.85 \text{ m} = 127.5 \text{ Nm}
    Moment due to girl = WG×1.2=490 N×1.2 m=588 NmW_G \times 1.2 = 490 \text{ N} \times 1.2 \text{ m} = 588 \text{ Nm}
    Anticlockwise moment:
    Moment due to father's support = RF×1.7R_F \times 1.7

    Sum of clockwise moments = Sum of anticlockwise moments

    487.5+127.5+588=RF×1.7487.5 + 127.5 + 588 = R_F \times 1.7 1203=1.7RF1203 = 1.7 R_F RF=12031.7R_F = \frac{1203}{1.7} RF=707.647 NR_F = 707.647 \text{ N}
  3. Apply the condition for translational equilibrium (sum of upward forces = sum of downward forces).

    Upward forces: RM+RFR_M + R_F
    Downward forces: WB+WP+WGW_B + W_P + W_G

    RM+RF=650+150+490R_M + R_F = 650 + 150 + 490 RM+RF=1290 NR_M + R_F = 1290 \text{ N}

    Substitute RFR_F:

    RM+707.647=1290R_M + 707.647 = 1290 RM=1290−707.647R_M = 1290 - 707.647 RM=582.353 NR_M = 582.353 \text{ N}
  4. State the reactions.

    The reaction of the mother is RMR_M and the reaction of the father is RFR_F.

    RM≈582.4 NR_M \approx \boxed{582.4 \text{ N}} RF≈707.6 NR_F \approx \boxed{707.6 \text{ N}}

    8. b) i.
    Relative motion describes the motion of an object with respect to another object or a specific frame of reference. It means that the observed velocity or position of an object depends on the motion of the observer.

    8. b) ii.
    Given: Velocity of car A, VA=100 ms−1V_A = 100 \text{ ms}^{-1}. Velocity of car B, VB=60 ms−1V_B = 60 \text{ ms}^{-1}.

    α. If the two cars are travelling in the same direction:
    The relative velocity of car A with respect to car B is the difference between their velocities.

    VAB=VA−VBV_{AB} = V_A - V_B VAB=100 ms−1−60 ms−1V_{AB} = 100 \text{ ms}^{-1} - 60 \text{ ms}^{-1} VAB=40 ms−1V_{AB} = \boxed{40 \text{ ms}^{-1}}

    β. If the two cars are travelling in opposite directions:
    The relative velocity of car A with respect to car B is the sum of their magnitudes of velocities.

    VAB=VA+VBV_{AB} = V_A + V_B VAB=100 ms−1+60 ms−1V_{AB} = 100 \text{ ms}^{-1} + 60 \text{ ms}^{-1} VAB=160 ms−1V_{AB} = \boxed{160 \text{ ms}^{-1}}

    8. c)
    Given: Mass m=4.0m = 4.0 kg. Initial velocity u=20 ms−1u = 20 \text{ ms}^{-1}. Final velocity v=30 ms−1v = 30 \text{ ms}^{-1}. Time t=5t = 5 s.

    i. Acceleration of the body:
    Using the equation of motion v=u+atv = u + at:

    30 ms−1=20 ms−1+a×5 s30 \text{ ms}^{-1} = 20 \text{ ms}^{-1} + a \times 5 \text{ s} 10 ms−1=a×5 s10 \text{ ms}^{-1} = a \times 5 \text{ s} a=10 ms−15 sa = \frac{10 \text{ ms}^{-1}}{5 \text{ s}} a=2 ms−2a = \boxed{2 \text{ ms}^{-2}}

    ii. Work done by the body:
    Work done is equal to the change in kinetic energy.

    W=ΔKE=12mv2−12mu2W = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 W=12m(v2−u2)W = \frac{1}{2}m(v^2 - u^2) W=12×4.0 kg×((30 ms−1)2−(20 ms−1)2)W = \frac{1}{2} \times 4.0 \text{ kg} \times ((30 \text{ ms}^{-1})^2 - (20 \text{ ms}^{-1})^2) W=2.0 kg×(900 m2s−2−400 m2s−2)W = 2.0 \text{ kg} \times (900 \text{ m}^2\text{s}^{-2} - 400 \text{ m}^2\text{s}^{-2}) W=2.0 kg×500 m2s−2W = 2.0 \text{ kg} \times 500 \text{ m}^2\text{s}^{-2} W=1000 JW = \boxed{1000 \text{ J}}

    iii. Change in momentum of the body:
    Change in momentum Δp=mv−mu\Delta p = mv - mu.

    Δp=m(v−u)\Delta p = m(v - u) Δp=4.0 kg×(30 ms−1−20 ms−1)\Delta p = 4.0 \text{ kg} \times (30 \text{ ms}^{-1} - 20 \text{ ms}^{-1}) Δp=4.0 kg×10 ms−1\Delta p = 4.0 \text{ kg} \times 10 \text{ ms}^{-1} Δp=40 kg ms−1\Delta p = \boxed{40 \text{ kg ms}^{-1}}

    9. a)
    Two factors that determine the quantity of heat in a body are:

    • The mass of the body.
    • The specific heat capacity of the material of the body.
    • The change in temperature of the body. (Any two of these are sufficient)

    9. b)
    Given: Mass of lead m=100m = 100 kg. Initial temperature T1=40 °CT_1 = 40 \text{ °C}. Melting point Tm=320 °CT_m = 320 \text{ °C}. Power of furnace P=10P = 10 kW =10×103= 10 \times 10^3 W. Specific heat capacity of lead c=120 Jkg−1K−1c = 120 \text{ Jkg}^{-1}\text{K}^{-1}. Specific latent heat of fusion of lead Lf=2.5×104 Jkg−1L_f = 2.5 \times 10^4 \text{ Jkg}^{-1}.

    i. Quantity of heat required to heat the lead to its melting point:
    The change in temperature ΔT=Tm−T1=320 °C−40 °C=280 °C\Delta T = T_m - T_1 = 320 \text{ °C} - 40 \text{ °C} = 280 \text{ °C}.
    Since 1 °C1 \text{ °C} change is equal to 1 K1 \text{ K} change, ΔT=280 K\Delta T = 280 \text{ K}.

    Q1=mcΔTQ_1 = mc\Delta T Q1=100 kg×120 Jkg−1K−1×280 KQ_1 = 100 \text{ kg} \times 120 \text{ Jkg}^{-1}\text{K}^{-1} \times 280 \text{ K} Q1=12000×280 JQ_1 = 12000 \times 280 \text{ J} Q1=3.36×106 JQ_1 = \boxed{3.36 \times 10^6 \text{ J}}

    ii. Additional heat required to melt the lead:
    This is the latent heat of fusion.

    Q2=mLfQ_2 = mL_f Q2=100 kg×2.5×104 Jkg−1Q_2 = 100 \text{ kg} \times 2.5 \times 10^4 \text{ Jkg}^{-1} Q2=250×104 JQ_2 = 250 \times 10^4 \text{ J} Q2=2.5×106 JQ_2 = \boxed{2.5 \times 10^6 \text{ J}}

    iii. Time taken to supply this additional energy:
    The power of the furnace is P=Q2tP = \frac{Q_2}{t}. We need to find tt.

    t=Q2Pt = \frac{Q_2}{P} t=2.5×106 J10×103 Wt = \frac{2.5 \times 10^6 \text{ J}}{10 \times 10^3 \text{ W}} t=2.5×106104 st = \frac{2.5 \times 10^6}{10^4} \text{ s} t=2.5×102 st = 2.5 \times 10^2 \text{ s} t=250 st = \boxed{250 \text{ s}}
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