Answer to Question #34051 in Mechanics | Relativity for abdullah ahsan

Question

A plank of mass 5 kg and length 3 meters is resting horizontally on two trestles P and Q which are a distance of 2.5 meter apart.When a student of mass 60 kg walks along the plank from one trestle to other, the plank sags.
Calculate the downward force exerted on each trestle when the student is a distance of 5 meters from the trestle

Answer 1

Let Fp and Fq bethe reaction forces at the trestle points assuming the plank is centered on the
trestle points from summing forces in the vertical direction to zero and
assuming up is positive we get:

Fp + Fq - (5 + 60)(10) = 0
Fp + Fq - 650 = 0
Fq = 650 - Fp

Summing moments about trestle support point q to zero and assuming a positive moment creates a CCW moment each moment will have aforce and a moment arm.

For student 0.5 m from p:

0 = 60(10)(2.0) + 5(10)(1.25) - Fp(2.5)
0 = 1200 + 62.5 - 2.5Fp
2.5Fp = 1262.5

Fp = 505 N

Fq = 650 - Fp
Fq = 650 - 505
Fq = 145 N

For student at 1.25 m from p which would be at the centerof the span between p and q summing moments about q to zero with same
assumptions as before:

0 = (60 + 5)(10)(1.25) - Fp(2.5)
0 = 812.5 - Fp(2.5)

Fp = 325 N

Fq = 650 - Fp
Fq = 650 - 325
Fq = 325 N