Calculate Force to Tip 5m Plank w/ 50kg Person

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Homework Help Overview

The discussion revolves around calculating the tipping point of a 5-meter uniform plank with a mass of 100 kg, which extends 2 meters over the edge of a building, when a 50 kg person walks on it. Participants are exploring the concepts of torque and balance of forces in this physics problem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the concept of torque and the balance of forces, questioning how to determine the point where the plank tips. There are attempts to clarify the role of the center of mass and the distances involved in the torque calculations.

Discussion Status

Some participants have provided guidance on identifying the pivot point and the distances relevant to the forces acting on the plank. There is an ongoing exploration of the calculations and interpretations of the distances involved, with no explicit consensus reached yet.

Contextual Notes

Participants are working within the constraints of the problem setup, including the mass of the plank and the person, as well as the geometry of the situation. There are indications of confusion regarding the distances and the implications of the plank's center of mass in relation to the edge of the building.

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A 5 meter uniform plank of mass 100kg rests on the top of a building with 2m extend over the edge of the building. how far can 50kg person can pass the edge of the building on the plank, before the plank begin to tip.

torque=Fd
i don't know where to get acceleration to calculate the force
 
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No need for an acceleration. Find the point where the torques are balanced. The weight of the plank acts as though it is applied at its center of mass. The distance is from the pivot point (edge of building) to the center of mass.
 
the point where torch is balance is 2.5m
i don't really understand what you say
 
The center of the plank is where the gravity acts on the mass. Since the plank is 5m, the center is 2.5m.

So you have one force.
 
Last edited:
So you have one force.
how so?
 
logglypop said:
the point where torch is balance is 2.5m
i don't really understand what you say

Draw a picture. Find the pivot point (point of rotation), it's the edge of the building. Find the distance from this point to the center of the plank (d1). Then call the distance from the pivot point to the person "d2." F1d1 =F2d2
 
mass of plank times acceleration due to gravity.
 
100(9.8)2.5=40(9.8)d
d=4.25?
 
logglypop said:
100(9.8)2.5=40(9.8)d
d=4.25?
Does that seem logical? a five meter plank, with 2 meters extending out over the edge of a building, and you can walk on it up to 4.5 meters beyond the edge of the building?

Draw it I said. The center of the plank is 2.5 m from the end of the plank, but how far is the center of the plank from the edge of the building, if 2 meters extends over the edge?

Draw it.Draw it. Go on, why don't you?
 
  • #10
i draw, i got .5 right?
 
  • #11
.5 meters, That's d1, right?
 
  • #12
100(9.8).5=50(9.8)d
d=1
 

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