Static Equilibrium of a beam and hinge

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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?
 

Answers and Replies

  • #2
1,196
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yah so i'm really not sure what I did wrong here
 
  • #3
1,196
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sorry i still dont get it
 
  • #4
PhanthomJay
Science Advisor
Homework Helper
Gold Member
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507
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.
 

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