Man on the end of a beam (rotational equilibrium problem)

In summary, the conversation discusses a problem involving a man standing on a uniform beam and the forces exerted on the beam by two supports. The person is trying to determine the distance the man is standing from the left support in order for the beam to be in equilibrium. The conversation includes a free body diagram and a discussion of using the torque equation to solve the problem. It is mentioned that the force exerted by the left-most support can be calculated, but it is not necessary to find the solution. The person also suggests defining up as positive and using this convention when calculating forces and torque.
  • #1
heatherwebb
1
0

Homework Statement


1. If a man weighing 800 Newtons is standing on a uniform beam an unknown distance from the supports. The beam is of length 5m and weighs 225 Newtons, and one support is positioned at the left end of the beam, the other is placed 1m from the left end and is exerting 600 Newtons up upon the beam. (http://heatherwebb.com/rotationalequilibriumdiagram.jpg - I did the parts in red)

a) Draw the free body diagram on the given picture, show all forces, known and unknown, along with all distances.

b) If the beam is in equilibrium, what distance is the man standing from the left support?



Homework Equations


Torque (left support) = Torque (man)


The Attempt at a Solution


Here is the diagram and what I've done to it, http://heatherwebb.com/rotationalequilibriumdiagram.jpg .
I know that the pivot exerts a force of 600N. Isn't that the normal force? Shouldn't I use that and the two known forces to determine the force at the left-most support? Then use that in the Torque equation?
It's not making sense to me because the force of the left-most support comes out to be negative.


The Attempt at a Solution


 
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  • #2
The drawing is correct. (The person could be at some place on the beam, not necessarily at the end).

I've worked out the force from the left-most support, and it is upwards. Maybe you mis-calculated? The best way to do it would be to define up as positive, so then all upward forces are positive and all downward forces are negative. Then all the forces added together should equal zero.

And yes, you're right that you need to use torque equation to see where the person must stand for there to be zero total torque. But actually, you don't need to know the force exerted by the left-most support to find this. Remember you can define the torque around any point, so what point will make this calculation easiest?
 

Related to Man on the end of a beam (rotational equilibrium problem)

What is the definition of rotational equilibrium?

Rotational equilibrium refers to the state in which a body or object is balanced and not rotating due to the sum of all the torques acting on it being equal to zero.

How do you determine the direction of a torque in a rotational equilibrium problem?

The direction of a torque in a rotational equilibrium problem can be determined using the right hand rule, where the fingers point in the direction of the force and the thumb points in the direction of the torque.

What are the key principles to keep in mind when solving a rotational equilibrium problem?

The key principles to keep in mind include: the sum of all forces acting on the body must be equal to zero, the sum of all torques acting on the body must be equal to zero, and the choice of pivot point can greatly affect the ease of solving the problem.

How do you choose a pivot point in a rotational equilibrium problem?

The pivot point should be chosen in a location where one of the forces acting on the body is zero, as this will eliminate the need to solve for its torque. It should also be chosen in a location that will result in the fewest number of unknown variables.

Can the concept of rotational equilibrium be applied to real-world situations?

Yes, the concept of rotational equilibrium can be applied to real-world situations such as balancing a seesaw or designing a bridge. It is an important principle in engineering and physics.

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