Question

Two forces F1 and F2 are lifting a heavy bucket vertically upward. The mass of the bucket is 7.5 kg. The magnitude of F2 = 88 N. F1 is 35 degrees from X- axis and F2 is 155 degrees from X-axis. Keep 2 decimal places in all answers. (a) What is the magnitude of the weight of the bucket in Newtons?
(b) Obtain an equation involving magnitudes of F1. Solve for F1?
(c) What is the y-component (in Newtons) of force vector F1?
(d) Find the acceleration ay in m/s^2?

Answer 1

Answer:

a) W= 73.5 N

b) F₁ = 97.36 N

c) F₁y= 55.84 N

d) ay = 2.60 m/s²

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in m/s²

Look at the attached graphic

Free body diagram of the bucket

α= 35° : angle that forms F₁ with the x axis

β = 180 -155= 25° :angle that forms F₂ with the x axis

F₂= 88N

m=7.5 kg : mass of the bucket

g= 9.8 m/s² : acceleration due to gravity

a) W: weight of the bucket

W = m*g = 7.5*9.8

W= 73.5 N

b)Equation involving magnitudes of F1. Solve for F1

∑Fx = m *ax          ax: acceleration in x , ax=0 ,

F₁* cosα- F₂*cosβ= 0

F₁* cos35° = F₂*cos25°

F₁ =  (88*cos25°)/( cos35°)

F₁ = 97.36 N

c) F₁y :  the y-component of force vector F₁

F₁y= 97.36* sin35°

F₁y= 55.84 N

d) ay : acceleration in y

∑Fy = m *ay    

F₁* sinα+F₂*sinβ -W= m *ay

97.36* sin35°°+88*sin25° -73.5= 7.5 *ay

19.53 = 7.5 *ay

ay = (19.53) / (7.5)

ay = 2.60 m/s²