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Define limiting friction and draw free body diagrams for a system with a 20 kg trolley on a surface with rolling friction and a 3 kg hanging block.

Physics

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Here are the solutions for questions 12.a and 12.b:

12.a) Limiting friction is the maximum static friction that can be exerted between two surfaces before one surface begins to slide over the other. It is the maximum force that opposes the initiation of motion.

12.b)
Free Body Diagrams (FBDs):

  • For the 20 kg trolley (W):

    • Weight (WTW_T): Acts vertically downwards. WT=Mg=20 kg×10 m/s2=200 NW_T = M g = 20 \text{ kg} \times 10 \text{ m/s}^2 = 200 \text{ N}.
    • Normal Force (NN): Acts vertically upwards, perpendicular to the surface. N=WT=200 NN = W_T = 200 \text{ N}.
    • Tension (TT): Acts horizontally to the right, due to the string.
    • Rolling Friction (frf_r): Acts horizontally to the left, opposing the motion. fr=μrNf_r = \mu_r N.
  • For the 3 kg hanging block:

    • Weight (WBW_B): Acts vertically downwards. WB=mg=3 kg×10 m/s2=30 NW_B = m g = 3 \text{ kg} \times 10 \text{ m/s}^2 = 30 \text{ N}.
    • Tension (TT): Acts vertically upwards, due to the string.

Acceleration of the system:

  1. Calculate the rolling friction force.

    The normal force on the trolley is N=MgN = M g.

    N=20 kg×10 m/s2=200 NN = 20 \text{ kg} \times 10 \text{ m/s}^2 = 200 \text{ N}

    The rolling friction force is fr=μrNf_r = \mu_r N.

    fr=0.04×200 N=8 Nf_r = 0.04 \times 200 \text{ N} = 8 \text{ N}
  2. Write down the equations of motion for each body using Newton's second law (F=maF=ma).

    For the 20 kg trolley (horizontal motion):

    T−fr=MaT - f_r = M a T−8 N=(20 kg)a(1)T - 8 \text{ N} = (20 \text{ kg}) a \quad (1)

    For the 3 kg hanging block (vertical motion, downwards is positive):

    mg−T=mam g - T = m a 30 N−T=(3 kg)a(2)30 \text{ N} - T = (3 \text{ kg}) a \quad (2)
  3. Solve the system of equations for acceleration (aa).

    Add equation (1) and equation (2):

    (T−8 N)+(30 N−T)=(20 kg)a+(3 kg)a(T - 8 \text{ N}) + (30 \text{ N} - T) = (20 \text{ kg}) a + (3 \text{ kg}) a 22 N=(23 kg)a22 \text{ N} = (23 \text{ kg}) a
  4. Calculate the acceleration.

    a=22 N23 kga = \frac{22 \text{ N}}{23 \text{ kg}} a≈0.9565 m/s2a \approx 0.9565 \text{ m/s}^2

    The acceleration of the block and the trolley system is 0.96 m/s2\boxed{\text{0.96 m/s}^2}.

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