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.

F = 4.8 N

α = 40.2°

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

there are no correct statementd on the list of choices you provided.

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