Monday

July 6, 2015

July 6, 2015

Posted by **Kate** on Tuesday, January 5, 2010 at 8:50pm.

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 9:52pmTake 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