A) The two conditions for a rigid body to be in equilibrium under a number of coplanar forces are:

Chemistry

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Answer

b

230 N\text{230 N}

170 N\text{170 N}

Step-by-step solution

5 steps

Here is the solution to the problem:

Part a

The two conditions for a rigid body to be in equilibrium under a number of coplanar forces are

  • The vector sum of all external forces acting on the body must be zero. This ensures translational equilibrium.
  • The vector sum of all external torques (or moments) about any point must be zero. This ensures rotational equilibrium.

Part b

230 N\text{230 N}

To determine the reactions at A and B, we apply the conditions for equilibrium.

  1. Identify all forces and their positions.

    • Weight of the uniform beam: W=100 NW = 100\, \text{N} acting at the center of the beam, which is at 5 m2=2.5 m\frac{5\, \text{m}}{2} = 2.5\, \text{m} from A.
    • Downward force: F1=200 NF_1 = 200\, \text{N} at 1 m1\, \text{m} from A.
    • Downward force: F2=100 NF_2 = 100\, \text{N} at 4 m4\, \text{m} from A.
    • Upward reaction force at A: RAR_A.
    • Upward reaction force at B: RBR_B. The total length of the beam is 5 m5\, \text{m}.

  2. Apply the condition for translational equilibrium (sum of vertical forces is zero).

    RA+RB−F1−W−F2=0R_A + R_B - F_1 - W - F_2 = 0 RA+RB−200 N−100 N−100 N=0R_A + R_B - 200\, \text{N} - 100\, \text{N} - 100\, \text{N} = 0 RA+RB=400 N(1)R_A + R_B = 400\, \text{N} \quad (1)

  3. Apply the condition for rotational equilibrium (sum of moments about any point is zero).

    Let's take moments about point A to find RBR_B.
    Clockwise moments = Anti-clockwise moments

    (F1×1 m)+(W×2.5 m)+(F2×4 m)=(RB×5 m)(F_1 \times 1\, \text{m}) + (W \times 2.5\, \text{m}) + (F_2 \times 4\, \text{m}) = (R_B \times 5\, \text{m}) (200 N×1 m)+(100 N×2.5 m)+(100 N×4 m)=(RB×5 m)(200\, \text{N} \times 1\, \text{m}) + (100\, \text{N} \times 2.5\, \text{m}) + (100\, \text{N} \times 4\, \text{m}) = (R_B \times 5\, \text{m}) 200 Nm+250 Nm+400 Nm=5RB m200\, \text{Nm} + 250\, \text{Nm} + 400\, \text{Nm} = 5 R_B\, \text{m} 850 Nm=5RB m850\, \text{Nm} = 5 R_B\, \text{m}

  4. Calculate RBR_B.

    RB=850 Nm5 mR_B = \frac{850\, \text{Nm}}{5\, \text{m}} RB=170 NR_B = 170\, \text{N}

  5. Substitute RBR_B into Equation (1) to find RAR_A.

    RA+170 N=400 NR_A + 170\, \text{N} = 400\, \text{N} RA=400 N−170 NR_A = 400\, \text{N} - 170\, \text{N} RA=230 NR_A = 230\, \text{N}

    The reactions at A and B are:
    RA=230 NR_A = \boxed{\text{230 N}}
    RB=170 NR_B = \boxed{\text{170 N}}

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