Engineering Mechanics - Force Vectors - Discussion
Discussion Forum : Force Vectors - General Questions (Q.No. 2)
2.
Determine the magnitude of the resultant force by adding the rectangular components of the three forces.
Discussion:
18 comments Page 1 of 2.
Des said:
1 year ago
Calculate the resultant of the three forces vector in Figure 1 by the component methods.
The magnitude of forces are F1=25.0N, F2=30.0N and F3 =40.0N.
The magnitude of forces are F1=25.0N, F2=30.0N and F3 =40.0N.
(1)
Ihuah patrick said:
3 years ago
f1x=-50cos30, f2x=75cos45, f3x = 50cos90.
f1y=50sin30, f2y=75sin45, f3y=-50sin90.
FRx = -50cos30 + 75cos45 + 50cos90 = 9.732.
FRy = 50sin30 + 75sin45 - 50sin90 = 28.033.
Magnitude of the resultant=square root of (sum of x components squared+sum of y components squared) = 29.7 N.
f1y=50sin30, f2y=75sin45, f3y=-50sin90.
FRx = -50cos30 + 75cos45 + 50cos90 = 9.732.
FRy = 50sin30 + 75sin45 - 50sin90 = 28.033.
Magnitude of the resultant=square root of (sum of x components squared+sum of y components squared) = 29.7 N.
(7)
Raymond said:
4 years ago
How did you get cos90?
(9)
J.BHARGAV SAI said:
4 years ago
Fx = - 50cos30 + 75cos45 + 50cos90
Fx = 9.73N.
Fy = 50sin30 + 75sin45 - 50sin90
Fy = 28.033N.
R = √{(Fx)^2 + (Fy)^2},
R = 29.723N.
Fx = 9.73N.
Fy = 50sin30 + 75sin45 - 50sin90
Fy = 28.033N.
R = √{(Fx)^2 + (Fy)^2},
R = 29.723N.
(4)
Pravin said:
4 years ago
Why can we not take the angle of F1 from the positive X-axis (150°) if I take 150° for F1 that would be wrong?
(2)
Seth said:
8 years ago
@Anand.
I am getting 43.3012709 how did you get that figure?
Please explain it.
I am getting 43.3012709 how did you get that figure?
Please explain it.
V Thimmaraju said:
10 years ago
I like this type of answers and I want many more problems and answers.
(1)
Prince tetteh said:
1 decade ago
The algebraic sum of vertical components of forces should be zero as well as horizontal forces.
Sum of vertical component = 75sin45+50sin30-50 = 28.033N.
Sum of horizontal component = 75cos45-50cos30 = 9.7317N.
Then magnitude of resultant = Root of the (Sum of vertical component squared+Sum of horizontal component squared) = 29.6741N.
Sum of vertical component = 75sin45+50sin30-50 = 28.033N.
Sum of horizontal component = 75cos45-50cos30 = 9.7317N.
Then magnitude of resultant = Root of the (Sum of vertical component squared+Sum of horizontal component squared) = 29.6741N.
(1)
Rajesh said:
1 decade ago
To resolve the all forces in cosine and sin components...like
R cos@=75cos45+50cos150+50cos270
R cos@=9.732
similarly
R sin@=28.037
Resultant is the square root of the sum of the square of the component...
R=29.7N
R cos@=75cos45+50cos150+50cos270
R cos@=9.732
similarly
R sin@=28.037
Resultant is the square root of the sum of the square of the component...
R=29.7N
(1)
Anand Reddi said:
1 decade ago
F2x=53.03
F2y=53.03
F1x=-43.3025
F1y=25
F in y direction=-50
By adding horizontal and vertical components then we get Resultant force F=9.7275i+28.03j
Magnitude is Sqrt((9.27^2)+(2803^2))==29.669936
F2y=53.03
F1x=-43.3025
F1y=25
F in y direction=-50
By adding horizontal and vertical components then we get Resultant force F=9.7275i+28.03j
Magnitude is Sqrt((9.27^2)+(2803^2))==29.669936
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