# Civil Engineering - Applied Mechanics - Discussion

Discussion Forum : Applied Mechanics - Section 1 (Q.No. 47)

47.

Two forces act an angle of 120°. If the greater force is 50 kg and their resultant is perpendicular to the smaller force, the smaller force is

Discussion:

9 comments Page 1 of 1.
Dawngtiltea said:
4 months ago

F1/sin150 = F2/sin90.

F1sin90 = 50sin150.

F1 = 150/sin90.

= 25.

F1sin90 = 50sin150.

F1 = 150/sin90.

= 25.

Amsal said:
4 years ago

50 * COS(180 - 120) = 25.

(2)

Mayur said:
6 years ago

I think it is solved by simple trignometry.

Gagan said:
6 years ago

120/50=2.4.

Then 180-120 = 60 then 60/2.4 = 25.

Then 180-120 = 60 then 60/2.4 = 25.

Hanan said:
6 years ago

Kg is not the unit for force.

Sanchit Mathur said:
6 years ago

Lami's Theorem says;

P/sinα = Q/sinβ = R/sinγ

In the above question, P=50, Q is the force perpendicular to smaller force and R is the smaller force;

α=90 , β=120, γ=360-(90+120).

This gives R as 25kg.

P/sinα = Q/sinβ = R/sinγ

In the above question, P=50, Q is the force perpendicular to smaller force and R is the smaller force;

α=90 , β=120, γ=360-(90+120).

This gives R as 25kg.

Subha Annadurai said:
7 years ago

How to use lami's theorem? Explain please.

Kishor said:
7 years ago

Here, we can also use Lami's theorem.

Abhay said:
8 years ago

Resultant force makes an angle α with the force P(in this case smaller force).

tanα = Qsinθ/(P+Qcosθ) α = 90, θ = 120, Q = 50, P =?

tanα = sinα/cosα = 1/0.

1/0 = Qsin(120)/(P+Qcos(120)) means P+Qcos(120) = 0.

Therefore P = -Qcos(120) = -50cos(120) = -50*(-1/2) = 25kg.

tanα = Qsinθ/(P+Qcosθ) α = 90, θ = 120, Q = 50, P =?

tanα = sinα/cosα = 1/0.

1/0 = Qsin(120)/(P+Qcos(120)) means P+Qcos(120) = 0.

Therefore P = -Qcos(120) = -50cos(120) = -50*(-1/2) = 25kg.

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