# Forces acting on a hinge!

1. Feb 2, 2010

### drkidd22

1. The problem statement, all variables and given/known data

In the figure below, the bar is 3.4 meters long and has a mass of 100 kg.
what is the magnitude of the hinge force?

3. The attempt at a solution

cosβ=2/sqrt(2^2+3.4^2)
cosβ=0.51
0.51T=mg/2
T= 100*9.8/(2*0.51)=960.8

I'm not 100% sure if I did this right.
Any assistance will be appreciated.

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2. Feb 2, 2010

### PhanthomJay

Looks like you took a short cut (not a good idea) to correctly solve for T, the cable tension force. But the problem wants the hinge force; try Newton 1 to solve for the x and y components of the force at the hinge.

3. Feb 2, 2010

### Spinnor

I think you got the tension right. Draw a free body diagram of the beam, label all the forces on the bar. The hinge applies a force on the beam in the x and y directions.

4. Feb 2, 2010

### drkidd22

well yeah, you are both right, but that's were I'm stuck.

5. Feb 2, 2010

### Spinnor

The beam has three forces acting on it. The tension force of the wire, the weight force of the beam and the force of the hinge which can be thought of as a sum of two forces, a force acting in the vertical direction on the beam and a force acting to the right on the beam.

Last edited: Feb 2, 2010
6. Feb 2, 2010

### drkidd22

so Fhx=Tsinβ= 489.95N?

7. Feb 2, 2010

### PhanthomJay

If cos B =0.51, then sin B =???

8. Feb 2, 2010

### Spinnor

Draw a picture roughly to scale. Fhx is considerably more then half T.

More like T*sin(60degrees)

9. Feb 2, 2010

### drkidd22

I'm completely lost. I give up. Will try again tomorrow

10. Feb 3, 2010

### drkidd22

= 59.34 degrees
sinB = 0.86

11. Feb 3, 2010

### PhanthomJay

Yes, that is correct. So if Fhx = T sinB, as you correctly noted, then Fhx =??? Now you still need to calculate Fhy.

12. Feb 3, 2010

### drkidd22

Fhx = 826.3N

13. Feb 3, 2010

### PhanthomJay

Yes, in which direction? And Fhy =??, and in which direction?? Once you get both components of the hinge force, the magnitude of the resultant hinge force is what the problem is asking.

14. Feb 3, 2010

### drkidd22

I think there is something wrong with the Tension. I still don't think I got it right.

15. Feb 3, 2010

### PhanthomJay

Except for some round off/sig figure errors, why do you think the tension is wrong? Your initial calculation for it is correct, although I question how you arrived at that formula.

16. Feb 3, 2010

### drkidd22

Well I think there is something wrong with the tension I found because I'm working on a similar problem with just a different length for the Bar and I use the same formula the results is not right. I get 1485N as the wire tension, but the book says the tension is 1095.

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17. Feb 3, 2010

### PhanthomJay

Don't believe everything you read, your answer is correct. But you should be summing moments and forces = 0 to calculate these values, do not blindly use formulas.

18. Feb 3, 2010

### drkidd22

19. Feb 3, 2010

### PhanthomJay

Are you familiar with the 3 basic equations of equilibrium (for objects at rest) which comes from Newton's 1st law:
Sum of all forces acting in the x direction = 0
Sum of all forces acting in the y direction = 0
Sum of all moments (torques) of forces about any point =0?
Break up the unknown force F (let's call it P instead of F, to avoid confusion of letter designations) into 2 components, P_x and P_y, where P_x = P cos30 and P_y = P sin30 (don't forget the direction of those force components). Now sum moments about the hinge end to solve for P_y; note that there is no moment from P_x, or from the wall forces, when you choose the hinge as your point of reference for determining moments. So for moments, you have the moment from P_y and the moment from the board weight (the board weight force acts at the center of the board). Add them up, set them equal to zero, and solve for P_y. Then you can get P_x and P from trig.