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A car is speeding up on a horizontal road with an acceleration a. Consider the following situations in the car. (1) A ball is suspended from the ceiling through a string and is maintaining a constant angle with the vertical. Find this angle. (2) A block is kept on a smooth incline and does not slip on the incline. Find the angle of the incline with the horizontal.


 A car is speeding up on a horizontal road with an acceleration a. Consider the following situations in the car. (1) A ball is suspended from the ceiling through a string and is maintaining a constant angle with the vertical. Find this angle. (2) A block is kept on a smooth incline and does not slip on the incline. Find the angle of the incline with the horizontal.


Grade:upto college level

1 Answers

Rohan
19 Points
6 years ago
-Forces on the ball hanging inside a speeding car As the forces on the ball are in equilibrium, equating the component of forces in vertical direction we get, T.cosθ = mg ----------(A) Equating the forces in horizontal direction we get, T.sinθ = ma ----------(B) Dividing (B) by (A), tanθ = a/g → θ = tan-1(a/g). Case (ii) Let the angle of the incline with the horizontal be θ, mass of the block "m" and the normal force on the block by the incline N. We want to apply Newton`s Laws of Motion with respect to the accelerating car which is a non-inertial frame. So additional pseudo force equal to `ma` will have to be considered acting on the block opposite to the direction of acceleration. Now forces on the block are weight mg downward, pseudo force ma opposite to `a` and the normal force N perpendicular to incline and upward as shown in the diagram below:-Forces on the block resting on the smooth incline inside a speeding carSince the block can move along the incline let us calculate the component of all the forces along the incline. Taking direction down the plane as positive, total force along the incline= mg.sinθ - ma.cosθ (Component of N is zero as Tcos90°=0) For the block not to slip on the incline this force should be zero. Thus, mg.sinθ - ma.cosθ = 0 g.sinθ = a.cosθ, →tanθ = a/g, → θ = tan-1(a/g).

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