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

The correct option is C $$m\sqrt { { g }^{ 2 }+{ a }^{ 2 } } +ma$$ It appears than three forces are applied to the bob1. Gravity2. Pseudo force (Due to acceleration of the car)3. Tension by the manBy 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 otherThe tension T not only has to balance this force but it also has to bring about an acceleration of a in the bobThus $$T-m(\sqrt{(g^2 + a^2)})= ma$$ (Newtons Laws , Net force = acceleration)$$T= m \sqrt{(g^2 + a^2)} + ma$$Physics

Suggest Corrections  0  Similar questions
View More  People also searched for
View More 