Posted by Kate on Tuesday, January 5, 2010 at 8:50pm.
[b]1. The problem statement, all variables and given/known data[/b]
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
[b]2. Relevant equations[/b]
in order to achieve static equilibrium
SIGMA torque = 0
SIGMA F_y = 0
SIGMA F_x = 0
[b]3. The attempt at a solution[/b]
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?

Physics  Damon, Tuesday, January 5, 2010 at 9:52pm
Take your torques about the hinge.
call the wire tension T
Calculate Ty, vertical component of tension.
55(.75) + 90(1.70)  .75 Ty = 0
(by the way cos40 =sin50 so we agree on equation )
so
Ty = 259 N
Sum vertical forces on hinge
F DOWN on hinge (is up on beam)  55 90 + 259 = 0
so F down on hinge = 114
so 114 N UP on hinge
The only horizontal force on the beam is
T cos 50
since T sin 50 = 259
T = 259/sin 50
and we want Tcos 50, thoe horizontal component
T cos 50 = 259 cos 50/sin 50
= 259/tan 50 = 217N toward the wall on the hinge
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