Engineering Mechanics - Force Vectors - Discussion

Solution:

First, we must add/ sum up all the forces for x-components. That is...

Fx(Rx)= F1x + F2x + F3x (since F3 is parallel to y-axis,F3x=0)
= -50cos30 + 75cos45 + 50cos90
Fx(Rx)= 9.7317...N

Then, add/sum up all the forces for y-components.

Fy(Ry)= F1y + F2y + F3y
= 50sin30 + 75sin45 - 50sin90
Fy(Ry)= 28.03300...N

R=*squareroot of [(Ry)^2 + (Rx)^2]
=*suareroot of [(28.03300859 N)^2 + (9.7317384 N)^2]
R=29.67416895 N

θ= arctan [Ry/Rx]
= arctan [28.03300859/9.7317384]
θ= 70.86 degrees

Therefore, the answer is A.

Get the x and y components of F1, F2 and F3
f1x = 50(cos 30)=43.301 ; f2x = 75 (cos 45) = 53.033
f1y = 50(sin 30)=25 ; f2y = 75 (sin 45) = 53.033
f3 = 50 (since f3 is parallel to y axis)

Summation along x = 53.033 - 43.301 = -22.426
Summation along y = 25 + 53.033 - 50 = 28.033

R = SQUARE ROOT OF (((22.426)square) + (((28.033)square))

R= 29.7

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.

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.

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

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

Fx=F1x+F2x+F3x=-50cos30+75cos45-50cos90=9.73n
Fy=F1y+F2y+F3y=50sin30+75sin45-50sin90=28.03n

Resultant force is,

R=square root of Fx^2+Fy^2.

You will get : 29.67.

R=29.67.

Fx = - 50cos30 + 75cos45 + 50cos90
Fx = 9.73N.

Fy = 50sin30 + 75sin45 - 50sin90
Fy = 28.033N.

R = √{(Fx)^2 + (Fy)^2},
R = 29.723N.

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.

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?