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 1 of 3.
ADITYA PRADHAN said:
1 decade ago
40*cos 120 = 40*1/2 = 20 N.
Raja said:
1 decade ago
Can anyone please explain?
Shahzad said:
10 years ago
40Cos 120° = 40*1/2 = 20 N.
Mohtashim Pathan said:
9 years ago
Smaller force = Bigger force * Cos (angle between two).
Mithilesh bhakta said:
9 years ago
40 * cos120 = 20N.
Md Naseeruddin said:
8 years ago
Tan (α) = psin (θ)/p + qcos (θ).
Alpha = 90, tan90 = 1/0.
θ = 120 = 180 - 120 = 60, sin60 = root3/2.
Cos60 = 1/2.
1/0 = p (root3/2)/p-40 (1/2).
P = 20.
Because cos (180-θ) = -cos (θ).
Alpha = 90, tan90 = 1/0.
θ = 120 = 180 - 120 = 60, sin60 = root3/2.
Cos60 = 1/2.
1/0 = p (root3/2)/p-40 (1/2).
P = 20.
Because cos (180-θ) = -cos (θ).
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.
Ravi said:
8 years ago
Here, Cos 120 is -1/2.
Abdul said:
7 years ago
It will be negative in direction but not by magnitude. As the force is in second quadrant where except sine value, all trigonometric values will be negative.
Sumit kumar said:
7 years ago
Q/sin30=P/sin90 by the fbd.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers