(^460) A Textbook of Engineering Mechanics
Example 22.2. Three links are hinged together to form a triangle ABC as shown in Fig. 22.4.
Fig. 22.4.
At a certain instant, the point A is moving towards the mid-point of BC with a velocity of 5 m/s,
and B is moving at a perpendicular direction to AC. Find the velocity of C.
Solution. Given : Velocity of A (vA) = 5 m/s
First of all, let us locate the position of instantaneous centre of the points A and B graphically
as shown in Fig. 22.5 and as discussed below :
- Draw the triangle ABC with the given data.
- Now draw the lines indicating the directions of
motions of points A (towards mid-point of BC) and
B (at right angles to AC). - Now draw perpendiculars at A and B on the
directions of motion of vA and vB. - Let these perpendiculars meet at O, which is the
instantaneous centre of the link AB and BC. - Now join OC and draw a line at right angle to OC
indicating the direction of motion of the point C.
Measuring the diagram to some scale, we find that OA = 2.6 cm and OC = 5.4 cm
We know that
5.4
2.08
2.6
C
A
v OC
vOA
===
∴ Velocity of C,
vC = vA × 2.08 = 5 × 2.08 = 10.4 m/s Ans.
EXERCISE 22.1
- The ends A and B of a link 1.5 m long are constrained to move in vertical and horizontal
guides as shown in Fig. 22.6.
At a given instant, when A is 0.9 m above C it was moving at 3 m/s upwards. Find the
velocity of B at this instant. (Ans. 2.25 m/s)
Fig. 22.6 Fig. 22.7
Fig. 22.5.