Resultant of 3 forces?
The resultant of the concurrent forces as shown is 1500N pointing up along the Y-axis. Compute the values of F and (phi) required to give this resultant..
in the diagram there are 3 forces one is 2500N
along the negative x axis two is F at (phi) on the first quadrant where (phi) is adjacent to the x-axis and 1200N at 30 deg. where the 30 deg. lies below the positive x axis..
i just don't know what method to use...
Answer:
Sum them together using vectors to find the resultant.
R = F1 + F2 + F3
we know that the resultant R is 1500N point up along the Y-axis so,
R = 0i + 1500j
call F1 = -2500i + 0j
F2 = F*cos(phi)i + F*sin(phi)j
F3 = 1200*cos(30) - 1200*sin(30)j
then R = (-2500+F*cos(phi)+1200*cos(30)... + (F*sin(phi) - 1200*sin(30))j
equating the vector terms,
the i and j terms
0 = -2500+F*cos(phi)+1200*cos(30)
1500 = F*sin(phi) - 1200*sin(30)
or
F*cos(phi) = 2500 - 1200*cos(30) = 1460.8N
F*sin(phi) = 1500 + 1200*sin(30) = 2100
using pythagorean theorem,
(F*cos(phi))^2 + (F*sin(phi))^2 = F^2
then
F = 2558.1 N
and
tan(phi) = (F*sin(phi))/(F*cos(phi)) = 1.4376
phi = 55.1774 degrees (could be 235.1774 but you said its in first quadrant...)
so F = 2558.1N and phi = 55.1774 deg
if u measure it right.
You need to use vector addition. Convert each force to a vector using the unit vectors along the coordinate axes. For example, the first force (2500N along the negative x axis) is represented in vector notation as [-2500, 0]. The second vector is [Fcos(phi), Fsin(phi)]), and the third vector is [1200cos(-30), 1200sin(-30)]. Add these three vectors to obtain:
[1200cos(-30) + Fcos(phi) - 2500, Fsin(phi) + 1200sin(-30)]
Set this result equal to the resultant shown ([0, 1500]), and you end up with a system of 2 equations in two unknowns (F and phi). Solve the system of equations for F and phi.
If you have a resultant of 2500N, three (x) component forces; F1(x)= -2500n, F2(x)= Fcos(phi), F3(x)= 1200N cos 30, and three (y) component forces; F1(y)= 0, F2(y)= Fsin(phi), F3(y)= -1200 sin30. Solve ΣF(x)= 0 because the resultant has only a y component. The forces in x must =0 so add F1(x) and F3(x) and solve for F2. F2=1460.8/ cosΦ. Same with ΣF(y)=2500N. Calculate F3(y) to get (2500+600)=F2 sinΦ. Substitute F2's to get 3100N=1460.8 (sinΦ/cosΦ). Solve for Φ by algebraic manipulation.
Since TanΦ=(sinΦ/cosΦ), then... (3100N/1460.8N)=tan Φ
therefore Φ=arctan(3100N/1460.8N)
Plug this Φ into prev equations to find F2.
I hope this explanation works without a graphic aid.
use simple lamis theorem and then apply conditon of reultant force
the sum of the forces vertical must be equal to zero and also the sume of the forces horizontal must equal zero. Therefore:
Sum of horizontal forces:
2500= Fcos phi + 1200 cos 30
2500 -1200(.866)= F cos phi
1126.67 = F cos phi
Sum of vertical forces:
1500+Fsin phi =1200 sin30
F sin phi =-900
solve the two equations for F and set equal to each other:
1126.67/cosphi = -900/sinphi
sin phi/cos phi = tan phi = -900/1126.67=-38.6184 deg
since this is negative the angle lies in the fourth quadrant and could be expressed as 321.3816 degrees
F sin phi = -900
F=-900/sin-38.6184
F=1131.833 N @ - 38.6184 deg or
F=1131.833 N @ 321.3816 deg
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