In another version of the \"Giant Swing\", the seat is connected to two cables a

Question

In another version of the "Giant Swing", the seat is connected to two cables as shown in the figure(Figure 1) , one of which is horizontal. The seat swings in a horizontal circle at a rate of 39.0 rev/min .

Part A

If the seat weighs 249 N and a 881-N person is sitting in it, find the tension in the horizontal cable.

Part B

If the seat weighs 249 N and a 881-N person is sitting in it, find the tension in the inclined cable.

1480

  N

" the second answer is correct but first keeps to appear wrong??"

1480

  N

Explanation / Answer

given

w = 39 rev/min

= 39*2*pi/60

= 4.08 rad/s

r = 7.5 m

m = 249/9.8 = 25.4 kg

M 881/9.8 = 89.9 kg

B)

In vertical direction the system is in equilibrium.

let Tv is the tension in the incined cable.
so, Fnety = 0

T*cos(40) - (m + M)*g = 0

T = (m + M)*g/cos(40)

= (249 + 881)/cos(40)

= 1475 N

A) let Th is the tension in the horizontal cable.

Apply, Fnetx = (m+M)*a_rad

Th + Tv*sin(40) = (m+M)*r*w^2

Th = (m+M)*r*w^2 - Tv*sin(40)

= (25.4 + 89.9)*7.5*4.08^2 - 1475*sin(40)

= 13447 N <<<<<<<<-------------------------------Answer

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