Calculate Force F at Point B for Slider along Guide

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To calculate the force F exerted by the guide on the slider at point B, the weight of the slider (0.12 kg) and its speed (1.4 m/s at point A) are used. The force at point A is calculated as mg, resulting in 1.177 N. At point B, the force differs due to the presence of normal acceleration, requiring the use of the formula F at B = Sqrt((mg)^2 + (ma)^2), where "a" represents normal acceleration. The solution involves recognizing that the speed at B is influenced by the curvature of the path, even if not explicitly stated. The discussion highlights the complexity of calculating forces in dynamic systems.
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If the slider along the smoth guide (i.e., no friction) on the horisontal plane has the speed 1.4m/s at A, what force F will the guide exert on the slider at point B? The weight of the slide is 0.12kg

http://img522.imageshack.us/img522/750/slider4lo.png

I'm able to calculate F at A = mg = 1.177. And the answer at B is different so I guess the speed should be different (but this is strange since it doesn't say anything about an acceleration in the tangential direction). But how do I get the speed at B? Thank you.
 
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From the drawing, it looks like the speed would be the same. Why would it be different?
 
How awfull isn't this? I've been working on this problem for hours now but my brain wasn't getting anywhere. While typing this message which was originally an explanation of my ideas I managed to come up with the solution.

Basically at A we have the force mg while we at B have two forces mg and ma ("a" here equals the normal acceleration). So F at B should be Sqrt((mg)^2+(ma)^2) = Sqrt((mg)^2+(mv^2/r)^2)

Anyway, thanks berkeman!
 
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