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
