A uniform 20 kg beam, 6 m long, rests on two supports; a 60 kg load hangs 2 m from the left one. Find the support reactions — Physics problems | PhysLabo
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Support reactions of a beam with a load

Medium 10 grade Mechanics / Statics
Problem statement

A uniform beam of mass $M = 20$ kg and length $L = 6$ m rests with its ends on two supports $A$ and $B$. A load of mass $m = 60$ kg hangs from the beam at a distance $a = 2$ m from support $A$ (see the figure). Find the support reactions $N_A$ and $N_B$. Take $g = 10$ m/s².

A uniform beam of length L on supports A and B with a load m hanging at distance a from A; weights and support reactions are shown
Given
  • $M = 20$ kg
  • $L = 6$ m
  • $m = 60$ kg
  • $a = 2$ m
  • $g = 10$ m/s²
  • $N_A = ?$
  • $N_B = ?$
Hints
Hint 1 / 2
The beam is in equilibrium: the net force is zero and the net torque about any point is zero. The weight of a uniform beam acts at its midpoint.
Hint 2 / 2
Take torques about support $A$ — then the unknown $N_A$ does not enter the equation. The lever arm of $N_B$ is $L$, of the load is $a$, of the beam weight is $L/2$.
Solution

The beam is in equilibrium: the net force is zero and the net torque about any axis is zero. The weight $Mg$ of the uniform beam acts at its midpoint.

Lever arms of the forces about support A
Lever arms of the forces about support A

1. Vertical forces:

$$N_A + N_B = Mg + mg = 200 + 600 = 800 \text{ N}$$

2. Torques about support $A$ (so $N_A$ drops out). The torque of $N_B$ turns the beam counterclockwise, the torques of the weights turn it clockwise:

$$N_B L - Mg\,\frac{L}{2} - mg\,a = 0$$$$N_B = \frac{Mg\,\frac{L}{2} + mg\,a}{L} = \frac{200 \cdot 3 + 600 \cdot 2}{6} = 300 \text{ N}$$

3. Then

$$N_A = 800 - N_B = 500 \text{ N}$$

Check with torques about $B$: $N_A L - mg(L - a) - Mg\,\frac{L}{2} = 3000 - 2400 - 600 = 0$ — correct.

Answer

$N_A = 500$ N, $N_B = 300$ N

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