(^32) A Textbook of Engineering Mechanics
∴
2
3
W
Pr×= ×r
π
or
2
3
W
P=
π
Ans.
Example 3.4. A uniform wheel of 600 mm diameter, weighing 5 kN rests against a rigid
rectangular block of 150 mm height as shown in Fig. 3.8.
Fig. 3.8.
Find the least pull, through the centre of the wheel, required just to turn the wheel over the
corner A of the block. Also find the reaction on the block. Take all the surfaces to be smooth.
Solution. Given : Diameter of wheel = 600 mm; Weight of wheel = 5 kN and height
of the block = 150 mm.
Least pull required just to turn the wheel over the corner.
Let P = Least pull required just to turn the wheel in kN.
A little consideration will show that for the least pull, it must be applied normal to AO. The
system of forces is shown in Fig. 3.9. From the geometry of the figure, we find that
150
sin 0.5
300
θ= = or θ = 30°
and AB==(300) – (150)^22 260 mm
Now taking moments about A and equating the same,
P × 300 = 5 × 260 = 1300
∴
1300
4.33 kN
300
P== Ans.
Reaction on the block
Let R = Reaction on the block in kN.
Resolving the forces horizontally and equating the same,
R cos 30° = P sin 30°
∴
sin 30° 4.33 0.5
2.5 kN
cos 30 0.866
P
R
×
== =
°^ Ans.
EXERCISE 3.1
- A rod AB 2·5 m long is supported at A and B. The rod is carrying a point load of 5 kN at
a distance of 1 m from A. What are the reactions at A and B ?[Ans. 2 kN ; 3 kN] - A beam AB of length 1 m is supported horizontally at A and B. A weight of 500 N is
attached on the rod at C. If the support at A can not bear a pressure more than 300 N, find
the distance of C from A, when the support A is about to fail. [Ans. 0.4 m]
Fig. 3.9.