Simple Conceptual Question about Forces.

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The discussion focuses on analyzing forces acting on a beam leaning against a frictionless wall and attached to a pin at the bottom. It clarifies that when considering a frictionless wall, no force acts parallel to the wall, which helps determine the direction of the contact force. The pin can exert both horizontal and vertical force components. The participant realizes that for equilibrium, the forces must balance, leading to a better understanding of the forces involved. This analysis is crucial for accurately drawing a free body diagram (FBD) of the beam.
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I am trying to draw an FBD for a beam leaning against a frictionless wall attached to a pin at the bottom end. The forces present would be the weight, normal force of the pin attached to the floor and the normal force of the wall.

When determining forces that are present, can I remove the object that is exerting the force on my primary object and then anaylze the logical motion that my primary object would follow?

For the case above, if I remove the frictionless wall, the beam would fall down and to the right? Wouldn't that mean there is a force that acts upwards and left? Same with the pin. If I removed the pin and floor, the bottom end of the beam would fall to the right and downwards, which would make me think that the force from the pin acts upwards and to the left.

My reasoning is wrong. So how would I analyze the forces at the pin and the wall?
 
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Your reasoning is wrong.

When you are told that the wall is frictionless, that means that there is no force exerted on the body parallel to the wall. Such a force would have to come from friction, and there is none. Therefore, you know the direction of the contact force at the wall.

At the pin, there can be force components in both directions, horiz & vert.

See if this wll get you going.
 
If the wall is frictionless, can there be a downward force acting on the beam at the wall? If the wall exerts a rightward horizontal force on the beam, what force must balance it for the equilibrium condition?
 
That makes sense. I knew there was something i was missing. Thanks alot.
 
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