Static Equilibrium of a beam and hinge

In summary, a 90 N sign is hanging on the end of a 55 N beam supported by a wire attached to the wall by a hinge. The horizontal and vertical components of the force on the hinge can be calculated using the equations for static equilibrium and summing moments about the hinge. After plugging in the values, the horizontal component is found to be 130 N and the vertical component is found to be -7.0 N. However, there may be an error in the calculation of the moment from the tension force, as it should be F_T_x times r3 instead of F_T_y times r3.
  • #1
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Homework Statement



A 90 N sign hangs on the end of a 55 N beam supported by a wire as shown. The beam is attached to the wall by a hinge. What are the horizontal and vertical componetns of the force on the hinge?

Notes:
the sign is 1.7 m from the wall
the CG of the beam is located .75 m from the wall and .95 m from the sign
at the CG of the beam the wire is attached with the other end attached to the wall
the angle that the beam makes with the wire is 50 degrees

Homework Equations



in order to achieve static equilibrium

SIGMA torque = 0
SIGMA F_y = 0
SIGMA F_x = 0

The Attempt at a Solution



subscripts
F_g is the force of gravity
F_h is the force of the hinge
F_T is the force of tension
_x was added on to forces to indicate a x component
_y was added on to forces to indicate a y component
_beam was added on to forces to indicate that a force exerted on the beam
_sign was added on to forces to indicated that a force exerted on the sign

SIGMA F_y = F_h_y + F_T_y - F_g_sign - F_g_beam = 0
SIGMA F_y = F_h_y + F_T sin THETA - F_g_sign - F_g_beam = 0

SIGMA F_x = F_h_x - F_T_x = 0
SIGMA F_x = F_h_x - F_T cos THETA= 0

PP at hinge
SIGMA torque = F_T_y * r_3 - F_g_beam * r_2 - F_g_sign * r_1 = 0
SIGMA torque = F_T sin THETA * r_3 - F_g_beam * r_2 - F_g_sign * r_1 = 0

add F_g_sign * r_1 to both sides
SIGMA torque = F_T sin THETA * r_3 - F_g_beam * r_2 = F_g_sign * r_1

add F_g_beam * r_2 to both sides
SIGMA torque = F_T sin THETA * r_3 = F_g_sign * r_1 + F_g_beam * r_2

divide both sides by sin THETA * r_3
SIGMA torque = F_T = ( F_g_sign * r_1 + F_g_beam * r_2 ) / sin THETA * r_3

plug and chug
SIGMA torque = F_T = ( 90 N(.95 m + .75 m) + 55 N * .75 m ) / (.75 m) sin 50
= 198.4 N

we know this
SIGMA F_x = F_h_x - F_T cos THETA= 0

add F_T cos THETA to both sides
SIGMA F_x = F_h_x = F_T cos THETA

plug chug
SIGMA F_x = F_h_x
= 198.4 N cos 50
= 130 N
rounded to two sig figs

we know this
SIGMA F_y = F_h_y + F_T sin THETA - F_g_sign - F_g_beam = 0

add F_g_sign to both sides
SIGMA F_y = F_h_y + F_T sin THETA - F_g_beam = F_g_sign

add F_g_beam to both sides
SIGMA F_y = F_h_y + F_T sin THETA = F_g_sign + F_g_beam

subtract F_T sin THETA from both sides
SIGMA F_y = F_h_y = F_g_sign + F_g_beam - F_T sin THETA

plug chug
SIGMA F_y = F_h_y = 90 N + 55 N - 198.4 N sin 50
= - 7.0 N

What gives? I should all of my steps... ALL of them... So could you please show me were I went wrong?
 
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  • #2
yah so I'm really not sure what I did wrong here
 
  • #3
sorry i still don't get it
 
  • #4
When you sum moments about the hinge, the moment from he tension force is F_T_x times r3, not F_T_y times r3, you got your sin and cos mixed up. I didn't check the rest of your work.
 

Related to Static Equilibrium of a beam and hinge

1. What is static equilibrium?

Static equilibrium refers to a state in which an object is at rest or moving at a constant velocity with no acceleration. This means that the forces acting on the object are balanced and the object is not experiencing any net force.

2. How does static equilibrium apply to a beam and hinge system?

In a beam and hinge system, static equilibrium means that the beam is at rest and not rotating. This requires the forces acting on the beam to be balanced and the sum of the moments around the hinge to be zero.

3. What is the role of the hinge in static equilibrium of a beam?

The hinge serves as a pivot point for the beam and allows it to rotate freely. It is also responsible for supporting the weight of the beam and any objects placed on it.

4. What factors affect the static equilibrium of a beam and hinge system?

The weight and distribution of the beam, the weight of any objects placed on the beam, and the location of the hinge all affect the static equilibrium of the system. Any external forces acting on the beam, such as wind or friction, can also impact the equilibrium.

5. How can I ensure static equilibrium in a beam and hinge system?

To ensure static equilibrium, you must make sure that the forces acting on the beam are balanced and the sum of the moments around the hinge is zero. This can be achieved by calculating and adjusting the weight and position of the beam and any objects placed on it.

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