# 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

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.

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.

Now, 60/6 = 10.

ANup said:
7 years ago

Equation moment about the pivot,

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.

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.

tan60 = 1.732.

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

Dinesh said:
8 years ago

Describe please?

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