Question

The 7.2 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 alpha with the horizontal. The 5.5 N force acts from above on the right at an angle of 48 degree with the horizontal. The force 7.2 N acts straight down. What is the magnitude of the force F? Answer in units of N. What is the angle alpha of the force F as shown in the figure above? Answer in units of degree.

Answer 1

F = 4.8 N

α = 40.2°

Step-by-step explanation:

Equilibrium of forces

∑Fx=0

5.5 cos48°- Fcosα = 0

5.5 cos48° = Fcosα

Fcosα = 3.68 Equation (1)

∑Fy=0

5.5 sin48°+ Fsinα - 7.2= 0

Fsinα = 7.2 - 5.5 sin48°

Fsinα = 3.11 Equation (2)

Equation (2) ÷ Equation (1)

Fsinα/ Fcosα= 3.11/ 3.68

sinα/ cosα= 3.11/ 3.68

tanα= 0.845

α = tan⁻¹ (0.845)

α = 40.2°

We replace α = 40.2° in the equation (1)

Fcosα = 3.68

Fcos(40.2)° = 3.68

F =3.68/cos(40.2)°

F = 4.8 N