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"The little ball with the mass of 100g has gotten stuck in a chute as depicted in the picture. What forces, and how large are they, that are acting on the ball?"


This is how I solve it:

  1. I find the force of gravity acting on the ball: $0.1kg * 9.8m/s = 0.98N$

  2. Next I construct a right triangle from the point of contact. Since the ball touches in two places the force of gravity is halved on each side.


Where $x$ is the normal force on the ball.

  1. $x = 0.49N * sin(20^o) = 0.17N$

  2. There's two of these normal forces, one on each side. As such we have a gravitational force of 0.98N and two normal forces, each of 0.17N.

The answer given in the book is that there is a gravitational force of 0.98N, and the normal force is 1.4N. What am I doing wrong here?

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You should decompose Normal force along the weight so that their horizontal component cancel out each other, not the weight along normal forces (as you end up with two weight component along two surfaces which won't cancel each other).

That is $F_n sin (20) = {W\over2}$ so $F_n = {W\over2 sin(20)} = 1.4 N$

Also in problems like these you should start by drawing all forces from a point and decomposing them according to one coordinate system.

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