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Question

A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hands of a person standing in the car. The car is moving with constant acceleration a directed horizontally as shown in $$Fig. 6.349$$. The other end of the string is pulled with constant acceleration a vertically. The tension in the string is equal to
981507_af2a18aaffd348efb3adf72da9dfa978.PNG


A
mg2+a2
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B
mg2+a2ma
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C
mg2+a2+ma
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D
m(g+a)
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Solution

The correct option is C $$m\sqrt { { g }^{ 2 }+{ a }^{ 2 } } +ma$$
433_28105_Untitled.jpg
It appears than three forces are applied to the bob

1. Gravity

2. Pseudo force (Due to acceleration of the car)

3. Tension by the man

By the law of addition of vectors the resultant of the two forces : the gravity and the pseudo force (The dotted line) = $$m(\sqrt{(g^2 + a^2)}) $$ as they are perpendicular to each other

The tension T not only has to balance this force but it also has to bring about an acceleration of a in the bob

Thus $$T-m(\sqrt{(g^2 + a^2)})= ma$$ (Newtons Laws , Net force = acceleration)

$$T= m \sqrt{(g^2 + a^2)} + ma$$

Physics

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