# Physics 5054 Doubts | Help Page 2

__Question 2: [Turning effect of forces]__
The diagram shows a
plank of mass 5.0 kg and length 3.00 m resting horizontally on two trestles, P
and Q, which are a distance of 2.50 m apart. When a student of mass 60.0 kg
walks along the plank from one trestle to the other, the plank sags.

**(a)**Explain why the sag increases as the student walks towards the middle of the plank.

**(b)**Calculate the downward force exerted on each trestle when the student is at a distance of

(i) 0.50 m from trestle
P,

(ii) 1.25 m from
trestle P.

(g = 10.0 N/kg)

**Reference:**

*Past Exam Paper –*

*N91 / P2 / QA2*

__Solution 2:__**(a)**

Assuming the plank is
uniform, the weight of the plank acts at the midpoint of the plank causing the
sag. As the bog walks from P towards the middle of the plank, the combined
centre of gravity of the system of plank and boy acts in between the boy and
the midpoint of the plank causing a larger sag. As the boy moves even nearer to
the midpoint, the sag increases as the combined c.g. moves towards the centre
of the plank.

**(b)**

(i) When x = 0.50 m from P,

By the Principle of
Moments, moment about P

R

_{Q}×2.50 = (600×0.50) + (50×1.25)
R

_{Q}= 145 N
For vertical equilibrium,

R

_{P}+ R_{Q }= 600 + 50 = 650 N
R

_{P}= 650 – R_{Q}= 650 – 145 = 505 N
(ii) When x = 1.25 m
from P, both the forces 600 N and 50 N act at the centre of mass of the plank.

By symmetry, R

_{P}= R_{Q}where R_{P}+ R_{Q}= 600 + 50 = 650 N
R

_{P}= R_{Q}= 650/2 = 325 N
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