A shop sign weighing 200 N hangs from the end of a uniform 160-N beam as shown i

Question

A shop sign weighing 200 N hangs from the end of a uniform 160-N beam as shown in (Figure 1) .

Part A

Find the tension in the supporting wire (at 35.0?).

Part B

Find the horizontal force exerted by the hinge on the beam at the wall.

Express your answer to three significant figures and include the appropriate units. Enter positive value if the direction of the force is to the right and negative value if the direction of the force is to the left.

Part C

Find the vertical force exerted by the hinge on the beam at the wall.

Explanation / Answer

A)

To find tension in the wire, we'll calculate the moment about the base of the beam:

weight of beam = 160 N acting at l=1.7m/2=0.85 m
weight of sign = 200 N acting at l=1.7m

only the vertical component of force in the wire will be used
because the horizontal component passes through the base

therefore Tvert=Tsin35 acting 1.35 m from the base

so here's the Moment eq

-200x1.7 (down) - 160x0.85 ( down ) + Tsin35 x 1.35 = 0
T=(200x1.7 + 160x0.85)/(1.35xsin35)
=614.72N (up and to the right )

B) from above the T_vert=Tsin35 = 352.58N (upwards)
and T_horiz=Tcos35=503.54 N (inward)

Now we can find vert and hor forces at the base

Horizontal====> horizontal Force + 503.54 N (inward ) = 0

horizontal force = -503.54 N (wall is pushing outward )
C)
Vert====>Sign + beam + T_vert + vert_base_force = 0
-200 (downward ) -160 (downward ) + 352.58(up) +vert_base_force=0
vert_base_force = 200 +160 - 352.58= 7.42 N (upward)

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