physics exam tmrw, really need help with this question! 10 pts!?

The way I solved it required no use of angles.

The centripetal force is the force holding in the object in question, which is the tension force's horizontal component.

Fc=Th=tension force, m=mass, r=radius

Fc=Th=mv^2/r

This will be equal to the horizontal component of the Tension

Vertical will be equal to the weight of the object.

Tv=mg

Since vertical, horizontal, and fullcomponents create a right triangle, use pythagrius therom to find the tension.

(mg)^2+(mv^2/r)^2=T^2 =. T=sqrt((mg)^2+(mv^2/r)^2)

The centripetal force needed to make something go in a circle is F = mv²/r

In this case F = 90x38²/17 = 7645N

This centripetal force is provided by the horizontal component of the tension.

Suppose the tension is T and the cable makes an angle θ to the vertical, then:

Tsinθ = 7645N

Weight = mg = 90 x 9.81 = 883N

There is no acceleration in a vertical direction, so the weight must be balanced by the vertical component of tension:

Tcosθ = 883

We can eliminate θ using the fact that sin²θ + cos²θ = 1

T²sin²θ = 7645²

T²cos²θ = 883²

T²sin²θ + T²cos²θ = 7645² + 883²

T²(sin²θ + cos²θ) = 7645² + 883²

T² = 7645² + 883²

T = √(7645² + 883²)

= 7700N to 2 significant figures

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