# Civil Engineering - Applied Mechanics - Discussion

Discussion Forum : Applied Mechanics - Section 1 (Q.No. 49)
49.
A uniform rod 9 m long weighing 40 kg is pivoted at a point 2 m from one end where a weight of 120 kg is suspended. The required force acting at the end in a direction perpendicular to rod to keep it equilibrium, at an inclination 60° with horizontal, is
40 kg
60 kg
10 kg
100 kg.
Explanation:
No answer description is available. Let's discuss.
Discussion:
8 comments Page 1 of 1.

Mess said:   2 years ago
c.g lies 4.5m which is at a distance of 4.5 - 2 = 2.5m from the pivot.

Ujjwal said:   4 years ago
Please, describe from where 2.5 cos 60 comes?

Golteb said:   5 years ago
Mb = 0.
120*2*cos60 - 40*2.5*cos60 -7Ra.
7Ra = 120-50.
Ra = 10kg.

Aiswarya said:   6 years ago
Can anyone please provide a pictorial representation for this question?

Sufazo said:   7 years ago
Simple 120 * 2/40 = 6.
Now, 60/6 = 10.

ANup said:   7 years ago

120 * 2cos(60) = 40 * 2.5 cos(60) + xcos(60)
or, 240 - 140 = 7x
x = 20kg.

Now, this is the weight of rod acting vertically downwards..to calculate the force perpendicular to the rod, resolve it in a direction perpendicular to the rod.

Hence,
Force = 20 cos (60) = 10kg.

Raj said:   7 years ago
40 * 9 * 2/120 = 720/120 = 6.

tan60 = 1.732.
tan60 * 6 = 1.732 * 6 = 10.39~= 10.

Dinesh said:   8 years ago