A particle, which remains at rest, is acted on by three forces F, G, and H.
The force F has magnitude 6N and is directed to the right and up, making an angle of 20' with the vertical.
The force G has magnitude 15N and is directed to the right and down, making an angle of 55' with the vertical.
Find the magnitude of the vector H, and the acute angle it makes with the vertical.
Make a reasonable accurate diagram.
Complete the parallelogram and draw in the resultant, you should be familiar with that.
The obtuse angle of the parallelogram is 105° and the acute angle is 75°
By the cosine law:
|r|^2 = 6^2 + 15^2 - 2(6)(15)cos75°
|r| = 14.6428...
angle made with F:
sin75/r = sinØ/15
sinØ = .98948..
Ø = appr 81.7° <--- angle between F and the resultant
but we want the equilibrant, which has opposite direction but the same magnitude.
I will leave it up to you to find the angle as requested.
Hint: Extend lines and use opposite angles.
After all, all angles are now known
To find the magnitude of vector H, we need to first resolve the forces F and G into their vertical and horizontal components. Then, we can use these components to find the net vertical and horizontal forces acting on the particle.
1. Resolving force F:
- The angle of 20' with the vertical means that it makes an angle of 90° - 20' = 89.67° with the horizontal.
- The horizontal component of F, Fx, can be found using cosine: Fx = F * cos(89.67°).
- The vertical component of F, Fy, can be found using sine: Fy = F * sin(89.67°).
2. Resolving force G:
- The angle of 55' with the vertical means that it makes an angle of 90° + 55' = 90.92° with the horizontal.
- The horizontal component of G, Gx, can be found using cosine: Gx = G * cos(90.92°).
- The vertical component of G, Gy, can be found using sine: Gy = G * sin(90.92°).
3. Finding the net horizontal and vertical forces:
- The horizontal component of the net force, F_netx, is the sum of Fx and Gx: F_netx = Fx + Gx.
- The vertical component of the net force, F_nety, is the sum of Fy and Gy: F_nety = Fy + Gy.
4. Finding the magnitude of vector H:
- Using Pythagoras' theorem, the magnitude of H, |H|, can be found: |H| = sqrt(F_netx^2 + F_nety^2).
5. Finding the acute angle with the vertical:
- The acute angle that vector H makes with the vertical, θ, can be found using the inverse tangent: θ = atan(F_nety / F_netx).
Now, let's plug in the values and calculate:
F = 6 N
G = 15 N
Angle F makes with the vertical = 20'
Angle G makes with the vertical = 55'
1. Resolving force F:
Fx = 6 * cos(89.67°) ≈ 0.034 N (horizontal component of F)
Fy = 6 * sin(89.67°) ≈ 5.996 N (vertical component of F)
2. Resolving force G:
Gx = 15 * cos(90.92°) ≈ -0.268 N (horizontal component of G)
Gy = 15 * sin(90.92°) ≈ 14.970 N (vertical component of G)
3. Finding the net horizontal and vertical forces:
F_netx = Fx + Gx ≈ 0.034 N + (-0.268 N) ≈ -0.234 N (horizontal component of the net force)
F_nety = Fy + Gy ≈ 5.996 N + 14.970 N ≈ 20.966 N (vertical component of the net force)
4. Finding the magnitude of vector H:
|H| = sqrt(F_netx^2 + F_nety^2) = sqrt((-0.234 N)^2 + (20.966 N)^2) ≈ 21.021 N