Question

Your friend's 10.8 g graduation tassel hangs on a string from his rearview mirror. When he accelerates from a stoplight, the tassel deflects backward toward the rear of the car at an angle of 5.23 ∘ relative to the vertical.
A) Find the tension in the string holding the tassel.
B) At what angle to the vertical will the tension in the string be twice the weight of the tassel?

Answer 1

The tension in the string holding the tassel and the vertical will the tension in the string

• T = 0.1953 N
• Ф = 34.4 °

What is the tension in the string holding the tassel. ?

Generally, the equation for Tension is  mathematically given as

$TCos\theta = mg$

Therefore

$TCos6.58^{o} = 19.8*10^{-3}*9.8$

T = 0.1953 N

b).

Where

$T* sin \theta = ma$

$0.1953*Sin6.58 \textdegree = 19.8*10^{-3}*a$

a = 1.13 m/s^2

In conclusion

T* sinФ = ma

2msinФ = ma

2sinФ = a

$sin\theta = \frac{a}{2}$

$\theta = sin^{-1}\frac{a}{2} \\\\\theta= sin^{-1}\frac{1.13}{2}$

Ф = 34.4 °

In conclusion, The tension in the string holding the tassel and the vertical will the tension in the string

T = 0.1953 N

Ф = 34.4 °

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