Question 1169864: A 0.7 kg ball attached to a string is rotating in a horizontal circle of radius 0.8m. The ball is attached to a line that dips 27 degrees below the horizontal. What is the tension in the string?
Answer: 15.11 N
and use phtagorean therome to solve, im stuck so could you help.. thanks!
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! Not sure about Pythagorean theorem and all that, but we can solve by
resolving the forces into horizontal and vertical components.
The tension, T, in the string is directed inward along the string.
In the y direction, there is a downward gravitational force equal to mg.
Since there is no acceleration in the y-direction, this must be balanced
by the y component of the tension, which is equal to Tsin(27).
Solving for T we get T = mg/sin(27) = 0.7*9.8/sin(27) = 15.11 N
There is a centripetal acceleration in the x-direction, which is
provided by the x-component of the tension, Tcos(27), but this is not required
to solve the problem.
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