Question

The 9.6 N weight is in equilibrium under the influence of the three forces acting on it. The F force acts from above on the left at an angle of α with the horizontal. The 6.9 N force acts from above on the right at an angle of 52◦ with the horizontal. The force 9.6 N acts straight down. What is the magnitude of the force F?

Answer 1

F= 5.95 N

Step-by-step explanation:

Forces acting on the body in equilibrium :

W = 9.6 N : in direcction vertical downward

F₁ = 6.9 N : acts from above on the right at an angle of 52° with the horizontal

F : acts from above on the left at an angle of α

x-y components of the force = 6.9 N

F₁x= 6.9 N*cos (52)°= 4.248 N

F₁y= 6.9N*sin (52)°= 5.437 N

Newton's first law for equilibrium

∑Fx=0

F₁x-Fx=0

Fx = F₁x

Fx = 4.248 N

∑Fy=0

F₁y + Fy -W = 0

Fy = W - F₁y

Fy = 9.6 N - 5.437 N

Fy = 4.16 N

Magnitude of the force F

`F=\sqrt{(F_(x))^(2) +(F_(y))^(2) }`

`F=\sqrt{(4.248)^(2) +( 4.16)^(2) }`

F= 5.95 N