Eugene E. answered • 12/02/15

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First identify the external forces acting on the seat. There is the tension T due to the rope, the gravitational force Fg (of magnitude 3g), and the pushing force, call it F. Since the seat is at rest, F(net,x) = 0 and F(net,y) = 0. We have

0 = F(net,x) = F sin(25) - T sin(55) (1)

0 = F(net,y) = F cos(25) + T cos(55) - 3g (2)

By (1),

F sin(25) = T sin(55) (3)

By (2),

3g - F cos(25) = T cos(25) (4)

Using (3) and (4), we write the proportion

F sin(25)/[3g - F cos(25)] = T sin(55)/[T cos(25)],

which is the same as

F sin(25)/[3g - F cos(25)] = tan(55) (5)

Multiplying (5) by 3g - F cos(25), we have

F sin(25) = 3g tan(55) - F cos(25)tan(55),

F sin(25) + F cos(25)tan(55) = 3g tan(55),

F[sin(25) + cos(25)tan(55)] = 3g tan(55).

Therefore

F = 3g tan(55)/[sin(25) + cos(25)tan(55)].

Since the pushing force makes an angle of 25 degrees with the upward vertical, the force vector is

-F sin(25) i + F cos(25) j = -{3g sin(25) tan(55)/[sin(25) + cos(25)tan(55)]} i + {3g cos(25) tan(55)/[sin(25) + cos(25)tan(55)]} j.

Eugene E.

tutor

Don't worry Ian, after doing more of these problems you'll feel more comfortable solving them. You can schedule an online lesson with me and we can work in more detail. :) Anyway, I've made some edits so hopefully it'll make more sense now. I misread the part about the angle of 25 being made with the upward vertical -- I thought it was the made with the horizontal! If you have more questions, please let me know.

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12/02/15

Ian G.

I keep confusing myself with the force diagram which i think i am getting wrong which is possibly confusing my sings

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12/02/15

Eugene E.

tutor

Hi Ian, send me an email so we can discuss further.

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12/03/15

Ian G.

12/02/15