College Physics

Figure 4.18When the vectors are projected onto vertical and horizontal axes, their components along those axes must add to zero, since the tightrope walker is

stationary. The small angle results inTbeing much greater thanw.

Consider the horizontal components of the forces (denoted with a subscriptx):

Fnetx=TLx−TRx. (4.43)

The net external horizontal forceFnetx= 0, since the person is stationary. Thus,

Fnetx= 0 = TLx−TRx (4.44)

TLx = TRx.

Now, observeFigure 4.18. You can use trigonometry to determine the magnitude ofTLandTR. Notice that:

(4.45)

cos (5.0º) =

TLx

TL

TLx = TLcos (5.0º)

cos (5.0º) =

TRx

TR

TRx = TRcos (5.0º).

EquatingTLxandTRx:

TLcos (5.0º) =TRcos (5.0º). (4.46)

Thus,

TL=TR=T, (4.47)

as predicted. Now, considering the vertical components (denoted by a subscripty), we can solve forT. Again, since the person is stationary,

Newton’s second law implies that netFy= 0. Thus, as illustrated in the free-body diagram inFigure 4.18,

Fnety=TLy+TRy−w= 0. (4.48)

ObservingFigure 4.18, we can use trigonometry to determine the relationship betweenTLy,TRy, andT. As we determined from the

analysis in the horizontal direction,TL=TR=T:

(4.49)

sin (5.0º) =

TLy

TL

TLy=TLsin (5.0º) = Tsin (5.0º)

sin (5.0º) =

TRy

TR

TRy=TRsin (5.0º) = Tsin (5.0º).

Now, we can substitute the values forTLyandTRy, into the net force equation in the vertical direction:

Fnety = TLy+TRy−w= 0 (4.50)

Fnety = Tsin (5.0º) +Tsin (5.0º) −w= 0

2 Tsin (5.0º) −w = 0

2 Tsin (5.0º) = w

and

142 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION

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