Tension in a string last problem.

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SUMMARY

The discussion centers on calculating the tension in a string of a pendulum bob weighing 1.0 kg, suspended from a 1.0 m long string at a 22-degree angle with the vertical. The tension in the string is determined to be 9.09 N using the formula Ft = mg cos(theta). The component of gravitational force acting perpendicular to the string is calculated as 3.67 N using Fg = mg sin(theta). Additionally, the acceleration of the pendulum bob is found to be 3.67 m/s². The discussion also addresses a query regarding the tension when the pendulum is held horizontally, emphasizing the need for a free body diagram (FBD) to resolve forces.

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Homework Statement



A 1.0 kg pendulum bob is suspended from a string 1.0 m long. The pendulum swings until the string makes an angle of 22 degrees with the vertical. At this point

Homework Equations



I can't remember

The Attempt at a Solution

a.) What is the tension in the string?

Answer: Ft = mgcostheta Ft = 9.09N

b.) What is the component of the force of gravity at right angles to the string?

Answer: Fg = mgsintheta = 3.67N

c.) What is the acceleration of the pendulum bob?

Answer: a = gsintheta = 3.67 m/s^2

d.) If the pendulum is held with a string fastened to the bob and held horizontally as shown, what would be the tension in the original string (not the horizontal one). (3 marks)

I don't know the answer to part D

The graph looks like this:

http://i53.tinypic.com/23rwa5l.jpg

Thanks.
 
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the tension would be in a way "shared" by both the strings. So drawing a FBD equate vertical and horizontal components and get your answer.
And your part 1) is wrong too,
Tcos(theta)=mg

And i think your c) part is wrong too, use the equation above, if it helps.
 

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