Thread: force question - urgent help please

Four forces Fsub1 , Fsub2, Fsub3, Fsub4 act on a particle in 2-space. The particle is in static equilibrium.
Fsub1 has magnitude 75N and points due east.
Fsub2 has magnitude 50N and points 30 degress north of east.
Fsub3 has magnitude 40N and points 60 degrees west of north.

Find the magnitude and direction of Fsub4.

Thank you very much.

Originally Posted by Jenny20

Four forces Fsub1 , Fsub2, Fsub3, Fsub4 act on a particle in 2-space. The particle is in static equilibrium.
Fsub1 has magnitude 75N and points due east.
Fsub2 has magnitude 50N and points 30 degress north of east.
Fsub3 has magnitude 40N and points 60 degrees west of north.

Find the magnitude and direction of Fsub4.

Thank you very much.

Express in components
(I understand that north of east means 30 degree above the east line)
(I understand that west of north means 60 degree to the left of north line).

Force 1)75 i
Force 2)50*cos(30)i+50*sin(30)j
Force 3)-40*cos(60)i+40*sin(60)
Force 4)F_xi+F_yj
Thus, sum of components,
75+50*cos(30)-40*cos(60)+F_x=0
50*sin(30)+40*sin(60)+F_y=0
You can solve.

Originally Posted by Jenny20

Four forces Fsub1 , Fsub2, Fsub3, Fsub4 act on a particle in 2-space. The particle is in static equilibrium.
Fsub1 has magnitude 75N and points due east.
Fsub2 has magnitude 50N and points 30 degress north of east.
Fsub3 has magnitude 40N and points 60 degrees west of north.

Find the magnitude and direction of Fsub4.

Thank you very much.

As F_1+F_2+F_3+F_4=0

F_4=-( F_1+F_2+F_3 )

Resolve the three forces you have into components in the N and E directions.
and sum the components of the three vectors, then F_4 is minus this sum.

You know how then to find the magnitude and direction of a vector in
component form.

RonL

I did not work out this question right in my homework. Can anyone show me the complete work of this question. Thank you very much!

Originally Posted by Jenny20

I did not work out this question right in my homework. Can anyone show me the complete work of this question. Thank you very much!

First do a diagram. See attachment.

We have:

f1=(75,0)
f2=(50cos(30),50sin(30))
f3=(-40sin(60),40cos(60))

so:

f4=-(75+50cos(30)-40sin(60), 50sin(30)+40cos(60))~=(83.66,45)

RonL

Originally Posted by Jenny20

Four forces Fsub1 , Fsub2, Fsub3, Fsub4 act on a particle in 2-space. The particle is in static equilibrium.
Fsub1 has magnitude 75N and points due east.
Fsub2 has magnitude 50N and points 30 degress north of east.
Fsub3 has magnitude 40N and points 60 degrees west of north.

Find the magnitude and direction of Fsub4.

Thank you very much.

Here is one way.

In a system in static equilibrium, the summation of forces is zero.
So F4, which is unknown, is the exact opposite of the Resultant of the 3 given forces.

You know how to get the Resultant?

Let the East-West axis be the x-axis.
And the North-South axis be the y-axis.

Rx = 75 +50cos(30deg) -40sin(60deg) = 83.66 newtons
Ry = 0 +50sin(30deg) +40cos(60deg) = 45 newtons

R = sqrt[(83.66)^2 +(45)^2] = 94.9947, or, 95 newtons
Angle from x-axis = arctan(45 /83.66) = 28.275 degrees north of East.

Therefore, F4 = 95 newtons, and points 28.275 degrees south of West. ----answer