Question

A 4 cm long rod of negligible weight is supported at a point 125 cm from its one end and a load of 18 kgf is suspended at a point 60 cm from the support on the shorter arm.
a) If a weight W is placed at a distance of 250 cm from the support on the longer arm, find W.
b) If a weight 5 kgf is kept to balance the rod find its position.
C



Ans:a) 4.32 kgf b) at a distance 2.16m from the support on longer arm

I need the solution. Answers are given.

Answer 1

A 4m long rod of negligible weight is supported at point 125cm from one end of it and a load of 18kgf suspended at a point 60cm from the support of shorter arm as shown in figure.
question said, a load of W kgf suspended at a point 250cm from the support of longer arm .

(A)
we know, rod will be horizontal only when ,
torque due to left load = torque due to right load
torque due to left load = 18kgf × (60/100)m
torque due to right load = W kgf × (250/100)m
now, 18kgf × 6/10 = W kgf × 25/10
W = 18 × 6/25 = 18 × 24/100
= 432/100 = 4.32kgf

(B) if W = 5kgf ,
Let at a distance x cm from the support of longer arm
now, 18kgf × 60cm = 5kgf × x cm
x = 18 × 60/5 = 18 × 12 = 216 cm
hence, at a distance 216cm from the support of longer arm

Answer 2

Given :
Load=L=18kgf
Load Arm=d=60cm
Effort=W=?
Effort arm=250cm

Let us use the principle of Lever
Effort x Effort arm=Loadxload arm

Wx250cm=18kgfx60cm
W=18x60/250
=1080/250
=4.32kgf

II) Load=18kgf
Load arm=d=60cm
effort=L=5kgf
Effort arm=D=?
By the principle of lever :

Effort x effort arm=load x load arm
5kgfxeffort arm=18kgfx60cm
Effort arm=18kgfx60cm/5kgf
=1080/5
=216cm=216/100=2.16m