- #1

userAA

## Homework Statement

Hi There,

I'm having trouble understanding (what I guess should be) a simple concept and so I can't apply it to a project I'm working on (also kind of rusty on torques and mechanics problems)

If I have a hinged beam (of uniform weight) and I lift it from one end and stay at rest, am I effectively lifting half the weight of the beam? (Please look at my calculation below). If so, how come when we lift the hood of our car it feels as though we are lifting less weight as the hood moves to a vertical position? Is it because it is rotational motion and I would have to consider some other equations (sorry I feel like I study this every year and forget it in a few days after not using it so I'm still referring to youtube videos for now).

The first two images are the question I have posted above (and my attempt at finding that the downward force being half the weight).

The last image is the mechanism I have to model the forces acting on point 2 that requires me to understand what force would be at the tip of the weight. (just so you can understand how this plays into why I'm trying to understand this). All lengths and angles are assumed to be known at every position.

Based on the mechanism I was also thinking that the reaction force applied from the uniform weight might be perpendicular to it but I couldn't convince myself why it would be.

Any direction would be greatly appreciated

## Homework Equations

Fx and Fy and Torque at equilibrium are equal to 0 (please have a look at my attempt at the solution)

## The Attempt at a Solution

∑Fx=0

Rx=0

∑Fy=0

Ry+Fperson-mg=0

∑T=0

@hinge

0=-mg(L/2)(cosθ)+Fperson(L)(cosθ)

simplifies to

Fperson=mg(0.5)

So force applied is half the weight

Problems on Google have the person applying a force but the vector is always shown to be perpendicular. In my case if you look at image 3 with the mechanism supporting the uniform weight, wouldn't the reaction force between the point it touches be exactly downwards?

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