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The answer according to solution manual: $F_A=566N F_B=283N$

I am not asking for you to solve this exercise, this is not a place for homework, I know. I need help how to think so I can solve this myself.

Clearly there will be one force downwards $mg=50\cdot9.8=490N$ and that will cause the ball to want to roll towards B. There will therefore be one force straight to left (the opposite force from B). Then I imagine that the force from A will go from A through the center of the ball and that this force times sin60=B more clearly $(F_A\cdot sin60=B$) but what will the force of A become and why? I saw someone dividing mg with cos30 to get A but why is this?

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closed as off-topic by Bill N, stafusa, Kyle Kanos, John Duffield, Jon Custer Mar 21 '18 at 13:59

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Bill N, stafusa, Kyle Kanos, John Duffield, Jon Custer
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Triple check your geometry and your trigonometry. There's a mistake. $\endgroup$ – Bill N Mar 20 '18 at 17:08
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From the diagram,

$Acos(60) = B$

$Asin(60) = Acos(30) = mg$

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