Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 5 (Q.No. 45)
45.
Two forces are acting at an angle of 120°. The bigger force is 40N and the resultant is perpendicular to the smaller one. The smaller force is
Discussion:
23 comments Page 2 of 3.
Gurunathan said:
6 years ago
tan@=1/0=Qsin 120/Q+40 cos 120.
By cross multiplying we get;
Q-20=0.
So Q= 20.
By cross multiplying we get;
Q-20=0.
So Q= 20.
Arifkhan said:
5 years ago
Cos@=bas/hyp.
Cos120=smaller-force /40,
And we get;
Smaller force = 20.
Cos120=smaller-force /40,
And we get;
Smaller force = 20.
Amit said:
8 years ago
2Pcosθ/2.
2 * 40 * (cos120)/ 2,
=40 * cos120,
=40 * 1/2.
= 20.
2 * 40 * (cos120)/ 2,
=40 * cos120,
=40 * 1/2.
= 20.
Khanka said:
6 years ago
If one angle is 120 then the other two must be less than 90.
Onkar kadam said:
5 years ago
Just use Lami' Theorem,
F/sin30 = 40/sin90,
F = 20 N.
F/sin30 = 40/sin90,
F = 20 N.
(5)
Mohtashim Pathan said:
9 years ago
Smaller force = Bigger force * Cos (angle between two).
Shahzad said:
10 years ago
40Cos 120° = 40*1/2 = 20 N.
Sumit kumar said:
7 years ago
Q/sin30=P/sin90 by the fbd.
ADITYA PRADHAN said:
1 decade ago
40*cos 120 = 40*1/2 = 20 N.
Raja said:
1 decade ago
Can anyone please explain?
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