Question

A wooden beam with a mass of 10.8 kg is supported by vertical supports

Answer 1

Answer:

Force from the support closest = 79.8 N

Force from the support furthest = 61.9 N

Explanation:

Let's say the length of the beam is L.  Let's say A is the near end of the beam and B is the far end of the beam.

Draw a free body diagram.  There are four forces on the beam:

Reaction force Ra at the near end (0),

Reaction force Rb at the far end (L),

Weight force of the beam Mg at the center (L/2),

Weight force of the book mg at L/4 from A.

Sum of torques at A:

∑τ = Iα

Rb (L) − Mg (L/2) − mg (L/4) = 0

Rb (L) = Mg (L/2) + mg (L/4)

Rb = ½ Mg + ¼ mg

Rb = (½ M + ¼ m) g

Rb = (½ (10.8 kg) + ¼ (3.66 kg)) (9.8 m/s²)

Rb = 61.9 N

Sum of forces in the y direction:

∑F = ma

Ra + Rb − Mg − mg = 0

Ra = Mg + mg − Rb

Ra = (M + m) g − Rb

Ra = (10.8 kg + 3.66 kg) (9.8 m/s²) − 61.9 N

Ra = 79.8 N