CK-12-Physics - Intermediate

8.4. Applications of Equilibrium Conditions http://www.ck12.org

Answer:The distance of the tank from ScaleBcan be expressed asL−x, where,x=vt. Thus,

FAL=mbg(L−x)+mpg

L

2

→mbg(L−vt)+mpg

L

2

→FA=

( 40 )( 3 − 0. 5 t)+ 60

( 3

2

)

3

→

FA= 40 −

20 t

3

+ 30 →FA= 70 −

20 t

3

Does the result satisfy the conditiont=0 in part 1b? How about when the books are on ScaleBat timet= 6 s?

http://demonstrations.wolfram.com/EquilibriumOfARigidBar/

Illustrative Example 4: A Sign of the Times

A sign of weight 100.0 N hangs from a rod of weight 80.0 N and length 1.2 m. A cable with unknown tensionTat
an angle of 60◦from the horizontal and a hingeHsupports the sign and rod. SeeFigure8.15 andFigure8.16.

FIGURE 8.15

Find the components of the forceFHthat the hinge places upon the rod.

Answer:

The equations for thexandyforces in translational equilibrium are (seeFigure8.16):

Equation 1:∑Fx= 0 →Hx−Tx= 0

Equation 2:∑Fy= 0 →Hy+Ty−Ws−Wr

We can also write the equation for rotational equilibrium:

Equation 3:∑τ= 0 →TyL−WrL 2 −WsL= 0

It will be more efficient to work with Equation 3 first since the only unknown in the equation isTy. Solving forTy,
we have∑τ= 0 →Ty= ( 80. 0 )

( 1. 20

2

)

+ 100. 0 ( 1. 20 ) = 168 N→Ty= 168 N.