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Center of gravity

  • Thread starter MAPgirl23
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  • #1
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A door of width 1.10 m and height 1.93 m weighs 290 N and is supported by two hinges, one a distance 0.500 m from the top and the other a distance 0.500 m from the bottom. Each hinge supports half the total weight of the door.

Assuming that the door's center of gravity is at its center, find the horizontal components of force exerted on the door by each hinge.

** Let H1 and H2 = forces exerted by the upper and lower hinges therefore:
H1_vert = H2_vert = weight/2 = 290 N / 2 = 145 N and the horizontal components of the hinge forces are equal in magnitude and opposite in direction.
net torque = 0 give H1_hor(1.10 m) - w(0.50 m) = 0 solve for H1_hor
H1_hor = 290N(0.50m/1.10m) - 145 N

I don't think this equation is correct.
 

Answers and Replies

  • #2
OlderDan
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MAPgirl23 said:
A door of width 1.10 m and height 1.93 m weighs 290 N and is supported by two hinges, one a distance 0.500 m from the top and the other a distance 0.500 m from the bottom. Each hinge supports half the total weight of the door.

Assuming that the door's center of gravity is at its center, find the horizontal components of force exerted on the door by each hinge.

** Let H1 and H2 = forces exerted by the upper and lower hinges therefore:
H1_vert = H2_vert = weight/2 = 290 N / 2 = 145 N and the horizontal components of the hinge forces are equal in magnitude and opposite in direction.
net torque = 0 give H1_hor(1.10 m) - w(0.50 m) = 0 solve for H1_hor
H1_hor = 290N(0.50m/1.10m) - 145 N

I don't think this equation is correct.
The torque equation is not correct. It appears you are calculating torque about hinge 2, and that is a good choice, but your distances are wrong. How far is hinge 1 from hinge 2? How far is the line of force of the weight from hinge 2?
 
  • #3
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Assuming that the door's center of gravity is at its center, the horizontal components of force exerted on the door by each hinge would be:
Let the doorweight be W.
Let the hinge's reactions be just like you described.

Moments wrt the lower hinge H2:
1.10/2*W-H1_hor*(1.93-2*0.5)=0
1.10/2*290-H1_hor*(1.93-2*0.5)=0

H1_hor=171.5 N

(1.10m/2)*290N - 171.5N(1.93m-1.00m) = 5 x 10^-3 is that correct?
 
  • #4
Pyrrhus
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If you the sum of moments at the hinge one

[tex] \sum \tau = H_{2_{x}}(1.93 - (0.5)(2)) - W_{door}(\frac{1.10}{2}) [/tex]
 
  • #5
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equates to 171.5N
 
  • #6
Pyrrhus
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Right, Now the sum of moments with respect to hinge 2

[tex] \sum \tau = -H_{1_{x}}(1.93 - (0.5)(2)) - W_{door} (\frac{1.10}{2}) [/tex]

The negative means the hinge 1 horizontal component must point left [itex] -x \vec{i} [/itex]

Hinge 1 for me was the top hinge, and hinge 2 the bottom hinge.
 
Last edited:
  • #7
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thanks, I understand
 
  • #8
Pyrrhus
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You know, what's interesting about this problem, it's that you could rewrite a equivalent system with a single couple, if you notice the vertical components on the hinges add up to the same magnitude of the weight of the door, so you could make the first couple, while the horizontal components of the hinges will give the 2nd couple, when both are added, we get our resultant couple.
 
  • #9
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it uses Newton's 3rd law
 

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