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:
10 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:
7 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:
7 years ago
Kg is not the unit for force.
Sanchit Mathur said:
7 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:
9 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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