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A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration a directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration a vertically. The tension in the string is equal to- (a) m sqrt(g^2 + a^2) (b) m sqrt(g^2 + a^2) -ma (c) m sqrt(g^2 + a^2) + ma (d) m (g + a) the answer is option c. can anyone tell me how to do it? please

A bob is hanging over a pulley inside a car through a string.
The second end of the string is in the hand of a person
standing in the car. The car is moving with constant
acceleration a directed horizontally as shown in figure. Other
end of the string is pulled with constant acceleration a
vertically. The tension in the string is equal to-
(a) m sqrt(g^2 + a^2) 
(b) m sqrt(g^2 + a^2) -ma
(c) m sqrt(g^2 + a^2) + ma 
(d) m (g + a)
the answer is option c. can anyone tell me how to do it? please

Grade:12

1 Answers

Swapnil Saxena
102 Points
9 years ago

433_28105_Untitled.jpg

Hi Goutham

The ques can be done like this.

First make a free body diagram of the entire arrangement

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(root(g^2 + a^2)) // as they are penpendicular to each other

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

Thus T-m(root(g^2 + a^2)) = ma (Newtons Laws , Net force = acceleration)

T= m root(g^2+a^2) + ma


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