(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

The 8kg slender bar is released from rest in the horizontal position. When it has fallen 45 degrees, what are the x and y components of force exerted on the bar by the pin support A?

2. Relevant equations

F=mawherea= at_hatt+ v^2/pn_hat

sin(45)=cos(45)=1/sqrt(2)

3. The attempt at a solution

The force exerted on bar by pin support is the inverted force acting on the bar downward. I see it as pretty much the same as a box on a floor, in order for the box not to go through the floor an equal force must be pushing it upward, so this should also be the case for the bar and pin support.

Thus I assume:F+mg.

In attached picture I show each force I know and from there I can see that these forces acting in each direction;

t_hat:ma+mg/sqrt(2)

a_hat:mv^2/p - mg/sqrt(2)

Now, the way I've put out my t-axis and n-axis I'll should get my invertedFas;F=-ma+mg/sqrt(2)t_hat+ mv^2/p - mg/sqrt(2)

This however is wrong. There is nothing else in this question I don't understand but this.

My professors answer on this is:F=ma-mg/sqrt(2)t_hat+ mv^2/p + mg/sqrt(2), and for some reason I just can't understand what he has done.

Can someone please explain why gravity acts negative int_hatand positive inn_hatwhile the rest ofFstays the same? I think I've wasted like 3 hours on this problem :P

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# Force exerted on bar by pin support

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