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
20 N
40 N
80 N
none of these
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
23 comments Page 2 of 3.

Farhad said:   6 years ago
Smaller/sin150 = bigger/sin90 = resultant/sin120.
= smaller/sin150 = bigger/sin90,
= smaller/sin30 = bigger/sin90,
= smaller = 0.5 * bigger.

Jitu said:   6 years ago
40 *sin30 = 20.

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.

Prashant said:   6 years ago
Simply take the components of force Q in x and Y direction.
Y direction gives magnitude of resultant and,
X direction gives magnitude of force P in opposite direction.

i.e. P - Qcosθ=0.
P = Q cosθ.
P = 40 cos60.
P = 20 N.

Khanka said:   6 years ago
If one angle is 120 then the other two must be less than 90.

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.

Rajdeep Singh said:   5 years ago
(120°-90°)=30°.
40sin(30°)=40*1/2=20N(force in opposite direction to smaller force).
Hence forec in smaller direction is - 20N.
Hence option D is the correct answer to this question.

Arifkhan said:   5 years ago
Cos@=bas/hyp.

Cos120=smaller-force /40,
And we get;
Smaller force = 20.

Rushi jadhav said:   5 years ago
40cos(120)

Cos(120) =-0.5=-1/2.
40*(-1/2) =-20.
-20 shows that it is in opposite directions.

Dibyendu Kundu said:   5 years ago
tan30°= Qsin120°/40+Qcos120°
0.577 = 0.866Q/40 + (-0.5Q),
23.08 - 0.28Q = 0.866Q,
23.08 = 0.866Q + 0.28Q,
23.08 = 1.146Q,
Q = 20.139 (approx 20N).


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