CK-12-Physics - Intermediate

http://www.ck12.org Chapter 8. Angular Motion and Statics

FA=

mtgxt+mbgL 2

L

=

(mtxt+mbL 2 )g

L

=

(

( 3. 6 × 104 kg)( 10 .0m)+( 6. 0 × 106 kg)

( 75 .0m

2

))

(10m/s^2 )

75 .0m

= 3 × 107 N

We solve forFBusing Equation 1.

FB=mtg+mbg−FA= (mt+mb)g−FA

= ( 3. 6 × 104 kg+ 6. 0 × 106 kg)(10m/s^2 )−( 3 × 107 N) = 3 .× 107 N

thusFB= 3 × 107 N.

Does it seem reasonable thatFAandFBshould be the same to only one significant figure? In the Check you
understanding section below, we do an example in which the support forces are very different.

Check Your Understanding

1a. Consider a similar example to Example 3 inFigure8.14, where a uniform plank with a massmpof 6.00-kg and
lengthLof 3.0 m is suspended from two bathroom scales,AandB. A set of books with massmb4.00 kg rests upon
the plank a distancexbof 0.75 m from ScaleA. Find the reading on each scale.

Choose your own pivot point and useg= 10 ms 2.

FIGURE 8.14

ForcesFAandFBhave the same magnitudes that scalesAandBread.

Answers:FA= 60 NandFB= 40 N(Hint: use the same equations as in Example 3.)

1b. If the books were sitting on the plank over ScaleA, what would each scale read?

Answer:ScaleAnow sees all of the weight of the books and ScaleBsees none of the weight. However, each scale
still supports half of the weight of the plank. Therefore ScaleAreadsFA= 40 N+ 30 N= 70 Nand ScaleBreads
FB= 0 N+ 30 N= 30 N.

1c. Here’s a challenge! Try this on your own before looking at the solution.

The books in the above example are substituted with a remote-controlled toy tank with the same mass as the books.

Assume the tank begins at ScaleAand moves with a constant velocity of 0.50 m/s from one end of the plank to the
other. Find an equation that gives the reading of ScaleAas a function of time.