Question #34051

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

Expert's answer

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

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

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