Awooden beam with a mass of 10.8 kg is supported by vertical supports
at each end. if a 3.66 kg textbook is placed halfway between one of the
supports and the middle of the wooden beam, what force must each
support provide if everything is to remain at rest?
(enter the force from the support closest to the book here)
(enter the force from the support farthest from the book here)

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

+15-35=-20

going in opposite directions

the journeys of marquette, joliet and la salle a) new france expand.

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