Three forces acting on an object are given F_1 = (-2.45i - 2.40j)N, F_2 = (5.40i

Question

Three forces acting on an object are given F_1 = (-2.45i - 2.40j)N, F_2 = (5.40i)N, and F_3 = (43.5i + 47.0j)N. The object experiences an acceleration of magnitude 3.50 m/s^2. What is the direction of the acceleration? What is the mass of the object? If the object is initially at rest, what is its speed after 18.0? What are the velocity components of the object after 18.0s? A bag of cement of weight 500 N hangs from three wires as suggested in the figure below. Two of the wires make angles theta_1 = 50.0 degrees and theta_2 = 27.0 degrees with the horizontal. Assuming the system is in equilibrium, Find the tensions in the wires.

Explanation / Answer

1.

Fnet = F1+F2+F3 = (-2.45 i -2.40j) + (5.40 i) + (43.5 i + 47.0 j) = (46.45i + 50.0j) N

a) =tan^-1(Fnety/Fnetx)= tan^-1(50.0/46.45) = 47.11 deg …above x axis

b) Fnet = sqrt(Fnetx^2+Fnety^2)= sqrt(46.45^2+50.0^2) = 68.25 N

m= Fnet/anet = 68.25/3.50 = 19.5 kg

c) vf= vi+at = 0 + 3.50*18.0 = 63.0 m/s

d) vfx = vix + anetx*t = 0 + 3.50cos47.11*18.0 = 42.88 m/s i

vfy = viy + anety*t = 0 + 3.50sin47.11*18.0 = 46.16 m/s j

vf = (42.88 m/s) i + (46.16 m/s) j

2.

Applying Newton’s second law at point B,

T3=mg = 500 N

Now applying Newton’s second law at point A horizontally,

T2cos27 - T1cos50 =0

T2= T1*(cos50/cos27) -------(1)

Now applying Newton’s second law at point A vertically,

T2sin27 + T1sin50 = T3----------(2)

From (1),

[T1*(cos50/cos27)]sin27 + T1sin50 = T3

T1*(cos50*tan27) + T1sin50 = T3

T1[(cos50*tan27) + sin50] = T3

T1[(cos50*tan27) + sin50] = 500

T1= 457.22 N

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