Engineering Mechanics - Force Vectors - Discussion

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.

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

@Anand.

I am getting 43.3012709 how did you get that figure?

Please explain it.

I get answer A. We divide the components in to horizontal and vertical components considering by respective axis.

Its an long method. Is there any shortcut of it.

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.

How come the answer get A. ?

I get D FOR the answer.

Please give me the solution.

I really need the solution.

How is it possible?