What is the tension in the cord holding a suspended sphere at a 23° angle?

AI Thread Summary
To find the tension in the cord holding a suspended sphere at a 23° angle, the vertical forces must be balanced since the sphere is in equilibrium. The gravitational force acting on the sphere is calculated using F_g = m*g, where the mass is 1.1 × 10^-4 kg and g is 9.8 m/s². The tension in the cord can be expressed as T = (mg) / sin(23°), resulting in a calculated value of approximately 27.58 × 10^-4 N. The user initially struggled with the calculations but recognized the need to apply trigonometric functions correctly to resolve the forces. The discussion emphasizes the importance of understanding equilibrium and the correct application of physics principles in problem-solving.
farleyknight
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Homework Statement



A sphere of mass 1.1 × 10^-4 kg is suspended from a cord. A steady horizontal breeze pushes the sphere so that the cord makes a constant angle of 23° with the vertical. Find (a) the magnitude of that push and (b) the tension in the cord.

Homework Equations



F_g = m*g

F_net = 0 (since we are in equilibrium)

The Attempt at a Solution



I am attempting (b) first, then (a) should follow naturally

Firstly, since the sphere is not moving, we have

F_net,y = 0

So then we can take the sum of the two forces acting along the vertical to get

F_g,y + T_y = 0

Where T_y is the vertical component of the tension in the cord. Taking the projection to the y coordinate, and expanding the definition of gravity, we have

-mg + T*sin(t) = 0

or

T = (mg) / sin(t) = (1.1*10^-4)*(9.8) / sin(23) == 27.58 * 10^-4 N

But this solution isn't working for me.. Any idea where I went wrong?
 
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T * cos23 = mg.
 
rl.bhat said:
T * cos23 = mg.

SOH-CAH-TOA, duh...

Thank you, sir
 
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