Question

In a simple jib crane, the vertical post is 2.4m high, the jib is 3.9m long and the tie is long. Find the forces on the jib and tie rod when a weight of 2.5kN is supported at the crane end.

Answer 1

Here it is given vertical post is 2.4 m high ,length of jib is 3.9 m long and tie is 2.1 m long

weight acting is 2.5 kN

since the line of action of all forces acts on a coincident point

so we can apply lami's theorem in it

and we can say that

$\frac{W}{sin \beta}=\frac{J}{sin \theta}=\frac{T}{sin \alpha}$

from above triangle we can find value of all sines

$sin \beta=\frac{2.4}{3.9}=0.615,sin \theta=1,sin\alpha=\frac{2.1}{3.9}$

Now we take one pair at time for finding values

so ,

$\frac{W}{sin \beta}=\frac{J}{sin \theta}$

$\frac{2.5}{0.615}=\frac{J}{1}$

J=4.065 kN (compressive load) this is the force on jib

$\frac{J}{sin \theta}=\frac{T}{sin \alpha}$

$\frac{4.065}{1}=\frac{T}{0.538}$

T=2.18697 kN (Tension) this is the force on tie