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.
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.
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.
M.Chandra Sekhar said:
4 years ago
By Formula,
Tan90=40Sin120/(P+40Cos120)
Since Tan90 = Infinitive..
The term 'P+40Cos120' must be equal to 0.
Therefore,
P + 40Cos120 = 0.
P = -40(-1/2); [ Cos120=Cos(90+30) = -Sin30 = -1/2],
P = 20N.
Tan90=40Sin120/(P+40Cos120)
Since Tan90 = Infinitive..
The term 'P+40Cos120' must be equal to 0.
Therefore,
P + 40Cos120 = 0.
P = -40(-1/2); [ Cos120=Cos(90+30) = -Sin30 = -1/2],
P = 20N.
(1)
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).
M.Chandra Sekhar said:
4 years ago
By Formula,
Tan90=40Sin120/(P+40Cos120)
Since Tan90=Infinitive.
The term 'P+40Cos120' must be equal to 0.
Therefore,
P+40Cos120=0
P = -40(-1/2); [ Cos120=Cos(90+30) = -Sin30 = -1/2]
P = 20N.
Tan90=40Sin120/(P+40Cos120)
Since Tan90=Infinitive.
The term 'P+40Cos120' must be equal to 0.
Therefore,
P+40Cos120=0
P = -40(-1/2); [ Cos120=Cos(90+30) = -Sin30 = -1/2]
P = 20N.
(3)
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.
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.
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).
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).
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.
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.
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.
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.
Cos(120) =-0.5=-1/2.
40*(-1/2) =-20.
-20 shows that it is in opposite directions.
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