"A pendulum consisting of an object of mass 1.3kg suspended from a string of length 1.9m is pulled aside so that the string makes an angle of 28 degrees with the vertical. At the instant the pendulum is released, what is the magnitude of the unbalanced force on the object?"

Question

I can't get this question, and am getting really frustrated. I know Fg is 12.74N, and should egual F1y. I know the angle between F1 and the horizontal is 62 degrees. But now Ihave no idea what to do. Oh, wait; I know F1x=cos62? So now I'm stuck. Please help?

Answer 1

Fg = (1.3kg)(9.8N/kg) = 12.74N
Fg can be resolved into two foces, T along the extension of the string, and Fp, at right angles to it.
Fp = (12.74N)(cos62)=________N, or
Fp = (12.74N)(sin28) =________N

The vector triangle I hope you drew, shows the weight, Fg resolved into two vectors, Fp, the unbalanced force perpendicular to the string, and T (string tension) which is along the extension of the string and cancelled out by the string resistance.

Answer 2

down load

Answer 3

6.78