Not sure if I have the correct angle and mass for this sum of forces

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Homework Help Overview

The discussion revolves around balancing forces in a system involving multiple masses and angles. The original poster presents equations for both east-west and north-south force balances, as well as moments, while attempting to determine the correct angle and mass for equilibrium.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to balance forces and moments, calculating angles and resultant forces based on trigonometric relationships. Some participants question the assumptions about the setup, including the orientation of the plate and the nature of the forces acting on it.

Discussion Status

Participants are exploring different interpretations of the problem setup and the forces involved. There is a lack of consensus on the correct approach, with some guidance offered regarding the assumptions about vertical and horizontal forces.

Contextual Notes

There is uncertainty regarding the orientation of the plate and the specific forces acting on it, as well as the distances represented in the diagram. Participants are questioning the appropriateness of balancing forces in the horizontal direction given the assumed vertical setup.

dougiehazard
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Homework Statement
Balance the following face plate if the masses are as follows:A = 2.8 kg, B = 3.5 kg, C = 4.2 kg, D = 4.8 kg, E = ?(What angle and mass should E be placed to balance the other 4 masses
angle.PNG

Balance the forces east-west:

3.5kg*sin45º + 4.2kg*cos30º - 4.8kg*sin30º - E*cosΘ = 0

E*cosΘ = 3.712 kg
balance north-south:

2.8kg + 3.5kg*cos45º - 4.2kg*sin30º - 4.8kg*cos30º + E*sinΘ = 0

E*sinΘ = 0.982
EsinΘ / EcosΘ = 0.982 / 3.712

tanΘ = 0.2645

Θ = 14.8º ◄
E = 3.712kg / cos14.8º = 3.84 kg ◄
If that's not right, then you want to balance the moments.
balancing east west:

3.5kg*0.35*sin45º + 4.2kg*0.315*cos30º - 4.8kg*0.6*sin30º - E*0.2*cosΘ = 0

E*cosΘ = 2.86 kg
and north-south

2.8kg*0.35 + 3.5kg*0.35*cos45º - 4.2kg*0.315*sin30º - 4.8kg*0.6*cos30º + E*0.2*sinΘ = 0

E*sinΘ = 6.55 kg
EsinΘ / EcosΘ = 6.55 / 2.86

tanΘ = 1.706

Θ = 59.6º ≈ 60º
E = 6.55kg / sin59.6º = 7.6 kg
 
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I assume you are asking which approach is correct.
Are we to understand that the plate is mounted vertically on a horizontal axis through the origin, and that the numbers in the diagram are distances of the masses from the origin?
What forces act where?
 
forces are the weight id assume and yes the number are distances in the diagram
 
dougiehazard said:
forces are the weight id assume
Then why are you considering a horizontal balance of forces?
Does any other force act vertically?