Download 11. To solve the problem, we note that acceleration is the second

11. To solve the problem, we note that acceleration is the second time derivative of the position
function; it is a vector and can be determined from its components. The net force is related to the
acceleration via Newton’s second law. Thus, differentiating
twice
with respect to t, we get
Similarly, differentiating
twice with respect to t yields
(a) The acceleration is
At
, we have
with a magnitude of
Thus, the magnitude of the force is
(b) The angle
or
We choose the latter (
makes with
) since
is
is in the third quadrant.
(c) The direction of travel is the direction of a tangent to the path, which is the direction of the
velocity vector:
At
makes with
, we have
is
We choose the latter (
Therefore, the angle
) since
is in the third quadrant.
13. (a) – (c) In all three cases the scale is not accelerating, which means that the two cords exert
forces of equal magnitude on it. The scale reads the magnitude of either of these forces. In each
case the tension force of the cord attached to the salami must be the same in magnitude as the
weight of the salami because the salami is not accelerating. Thus the scale reading is mg, where
m is the mass of the salami. Its value is (11.0 kg) (9.8 m/s2) = 108 N.
15. (a) From the fact that T3 = 9.8 N, we conclude the mass of disk D is 1.0 kg. Both this and
that of disk C cause the tension T2 = 49 N, which allows us to conclude that disk C has a mass of
4.0 kg. The weights of these two disks plus that of disk B determine the tension T1 = 58.8 N,
which leads to the conclusion that mB = 1.0 kg. The weights of all the disks must add to the 98 N
force described in the problem; therefore, disk A has mass 4.0 kg.
(b) mB = 1.0 kg, as found in part (a).
(c) mC = 4.0 kg, as found in part (a).
(d) mD = 1.0 kg, as found in part (a).
57. We take +y to be up for both the monkey and the package.
!"#$%&$'()&$*!$'+$,&$-.$/+0$,+'1$'1&$2+3)&4$(35$'1&$.(6)(7&#$
(a) The force the monkey pulls downward on the rope has magnitude F. According to Newton’s
8(9$ :1&$ /+06&$ '1&$ 2+3)&4$ .-;;<$ 5+=3=(05$ +3$ '1&$ 0+.&$ 1(<$ 2(73>'-5&$"#$ ?66+05>37$ '+$
third law, the rope pulls upward on the monkey with a force of the same magnitude, so Newton’s
@&='+3A<$ '1>05$ ;(=B$ '1&$ 0+.&$ .-;;<$ -.=(05$ +3$ '1&$ 2+3)&4$ =>'1$ ($ /+06&$ +/$ '1&$ <(2&$
second law for forces acting on the monkey leads to
2(73>'-5&B$<+$@&='+3A<$<&6+35$;(=$/+0$/+06&<$(6'>37$+3$'1&$2+3)&4$;&(5<$'+
F – mmg = mmam,
"$C$##$$D$#
#%#B
where #
mm$><$'1&$2(<<$+/$'1&$2+3)&4$(35$%
is the mass of the monkey and am $><$>'<$(66&;&0('>+3#$E>36&$'1&$0+.&$><$2(<<;&<<$
is its acceleration. Since the rope is massless F = T is
=1&0&
#
#
the
tension
in
the
rope.
The
rope
pulls
upward
on the package with a force of magnitude F, so
"$D$&$><$'1&$'&3<>+3$>3$'1&$0+.&#$:1&$0+.&$.-;;<$-.=(05$+3$'1&$.(6)(7&$=>'1$($/+06&$+/$
Newton’s second law for the package is
2(73>'-5&$"B$<+$@&='+3A<$<&6+35$;(=$/+0$'1&$.(6)(7&$><
F +$C$#
FN –$$D$#
mpg =% mB pap,
"$*$"
'
(
( (
where #
mp$ is
the mass of the package, ap is its acceleration, and FN is the normal force exerted by
=1&0&
( ><$ '1&$ 2(<<$ +/$ '1&$ .(6)(7&B$ %($ ><$ >'<$ (66&;&0('>+3B$ (35$ "'$ ><$ '1&$ 3+02(;$ /+06&$
the ground on it. Now, if F is the minimum force required to lift the package, then FN = 0 and ap
&F&0'&5$,4$'1&$70+-35$+3$>'#$@+=B$>/$"$><$'1&$2>3>2-2$/+06&$0&G->0&5$'+$;>/'$'1&$.(6)(7&B$
= 0. According
to %&%'*%+,,-.)/$0%!-%!"#%1#,-$)%2(3%#45(!/-$%6-.%!"#%7(,8(0#9%!"/1%:#($1%
the second law equation for the package, this means F = mpg. Substituting mpg
!"#$
!"%&%'%($)%#
$
for F in&*%;5<1!/!5!/$0%%
the equation for &%6-.%!%/$%!"#%#45(!/-$%6-.%!"#%:-$8#=9%3#%1-2>#%6-.%#
the monkey, we solve for am:
!%&%%
$
$
%?
@
! − %% & ( % $ − %% ) & (AB 80 − A' 80 ) ( C*D :E1 )
=
=
= F*C :E1 @ *
#% =
%%
%%
A' 80
(b) +1%
As )/1,511#)9%
discussed,I#3!-$J1%
Newton’s1#,-$)%
second
leads
for7(,8(0#%
the package
G<H%
2(3%law
2#()1%
!-% !to− % $ & = % $ # $ %6-.% !"#%
($)% and
for!"#%
the:-$8#=*%
monkey.K6%If!"#%
the(,,#2#.(!/-$%
acceleration-6%of
package
is downward,
then the
! − %% & = %% #% %6-.%
!"#%the
7(,8(0#%
/1% )-3$3(.)9%
!"#$%
acceleration of the monkey is upward, so am = %–a
the first equation for F
p. Solving
%&%L#
!"#%(,,#2#.(!/-$%-6%!"#%:-$8#=%/1%573(.)9%1-%#
$*%;-2>/$0%!"#%6/.1!%#45(!/-$%6-.%!
!
"
#
! = %$ & + # $ = %$ & − #%
$
($)%15<1!/!5!/$0%!"/1%.#152!%/$!-%!"#%1#,-$)%#45(!/-$9%3#%1-2>#%6-.%#%?
#
(%
=
$
− %% ) &
=
(AB 80 − A' 80 ) ( C*D :E1@ )
= @*' :E1 @ *
and substituting this result into the second equation, we solve for am:
(c) The result is positive, indicating that the acceleration of the monkey is upward.
(d) Solving the second law equation for the package, we obtain
60. The motion of the man-and-chair is positive if upward.
(a) When the man is grasping the rope, pulling with a force equal to the tension T in the rope, the
total upward force on the man-and-chair due its two contact points with the rope is 2T. Thus,
Newton’s second law leads to
so that when a = 0, the tension is T = 466 N.
(b) When a = +1.30 m/s2 the equation in part (a) predicts that the tension will be
.
(c) When the man is not holding the rope (instead, the co-worker attached to the ground is
pulling on the rope with a force equal to the tension T in it), there is only one contact point
between the rope and the man-and-chair, and Newton’s second law now leads to
so that when a = 0, the tension is T = 931 N.
(d) When a = +1.30 m/s2, the equation in (c) yields T = 1.05 × 103 N.
(e) The rope comes into contact (pulling down in each case) at the left edge and the right edge of
the pulley, producing a total downward force of magnitude 2T on the ceiling. Thus, in part (a)
this gives 2T = 931 N.
(f) In part (b) the downward force on the ceiling has magnitude 2T = 1.05 × 103 N.
(g) In part (c) the downward force on the ceiling has magnitude 2T = 1.86 × 103 N.
(h) In part (d) the downward force on the ceiling has magnitude 2T = 2.11 × 103 N.
61. The forces on the balloon are the force of gravity
(down) and the force of the air
(up). We take the +y to be up, and use a to mean the magnitude of the acceleration (which is not
G
!"#$%&'$()*+',$)-$.&'$/011))-$0*'$.&'$()*+'$)($2*034.5$ !" $67)8-9$0-7$.&'$()*+'$)($.&'$04*$
G
#$ $6:;9#$<'$.0='$.&'$>%$.)$/'$:;?$0-7$:,'$ $$.)$@'0-$.&'$ !$"&'()*+$)($.&'$0++'1'*0.4)-$
its 68&4+&$
usual use
in 4.,$
this:,:01$
chapter).
When
the mass<&'-$
is M .&'$
(before
is .&'$
thrown
out)
4,$ -).$
:,'$ 4-$
.&4,$ +&0;.'*9#$
@0,,$the
4,$ ,ballast
$ 6/'()*'$
/0110,.$
4,$ the
acceleration
is downward and Newton’s second law is
.&*)8-$):.9$.&'$0++'1'*0.4)-$4,$7)8-80*7$0-7$A'8.)-B,$,'+)-7$108$4,$$
Mg$D$C
= –Ma.
#F$a$C$–,"
,$#
After
the ballast is thrown out, the mass is M,–$C$m!(where
is the mass of the ballast) and the
E(.'*$.&'$/0110,.$4,$.&*)8-$):.?$.&'$@0,,$4,$
$68&'*'$m!$4,$.&'$@0,,$)($.&'$/0110,.9$0-7$
acceleration
is
upward.
Newton’s
second
law
leads
to
.&'$0++'1'*0.4)-$4,$:;80*7#$A'8.)-B,$,'+)-7$108$1'07,$.)
– (M
= (M
#F$a$C$6
,$C$–!m)g
9"$D$6
,$C$–!m)a.
9$#
The%&'$;*'34):,$'F:0.4)-$243',$
previous equation gives Fa =
this plugs into the new equation to give
#$M(g
$D$,–6"a),
$C$$and
9?$0-7$.&4,$;1:2,$4-.)$.&'$-'8$'F:0.4)-$.)$243'$
!
" !
" !
"
, " −$ − , −! " = ,−!$ Ÿ !=
G ,$
#
"+$