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.
Shishir Biswas said:
6 years ago
R makes angle with 40 is 30°
So, tan30°=P sin120°/40+P cos120°
Solve and find P = 20 N.
So, tan30°=P sin120°/40+P cos120°
Solve and find P = 20 N.
Jitu said:
6 years ago
40 *sin30 = 20.
Farhad said:
6 years ago
Smaller/sin150 = bigger/sin90 = resultant/sin120.
= smaller/sin150 = bigger/sin90,
= smaller/sin30 = bigger/sin90,
= smaller = 0.5 * bigger.
= smaller/sin150 = bigger/sin90,
= smaller/sin30 = bigger/sin90,
= smaller = 0.5 * bigger.
Sumit kumar said:
7 years ago
Q/sin30=P/sin90 by the fbd.
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.
Ravi said:
8 years ago
Here, Cos 120 is -1/2.
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.
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).
Mithilesh bhakta said:
9 years ago
40 * cos120 = 20N.
Mohtashim Pathan said:
9 years ago
Smaller force = Bigger force * Cos (angle between two).
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