𝑎Ԧ p when x & & & & & & 160 & & v v i v j a a i a j -...
TRANSCRIPT
7. For a certain interval of motion, the pin P is forced to move in the fixed parabolic
slot by the vertical slotted guide, which moves in the x direction at the constant rate
of 40 mm/s. All measurements are in mm and s. Calculate the magnitudes of Ԧ𝑣 and
Ԧ𝑎 of pin P when x = 60 mm.
SOLUTION
Trajectory of the pin is160
2xy ??60 avmmxwhen
jaiaajvivv yxyx
jaiaajvivv yxyx
yaxayvxv yxyx ,,
0)(/40 xacstsmmxv xx
The first derivative of trajectory equation with respect to time
smm
xxxxy
xy /30
80
4060
80160
2
160
2
jiv
3040 Magnitude of velocity of the pin: smmv /503040 22
22
2 /2080
40
80
1
80smmxxxy
xx
dt
dy
dt
d
ja
20
Magnitude of acceleration of the pin:
2/20 smma
8. Pins A and B must always remain in the vertical slot of yoke C, which
moves to the right at a constant speed of 6 cm/s. Furthermore, the pins
cannot leave the elliptic slot. What is the speed at which the pins approach
each other when the yoke slot is at x = 50 cm? What is the rate of change
of speed toward each other when the yoke slot is again at x = 50 cm?
100 cm
60 cm
x
6 cm/s
yoke
Cx
y
10. A long-range artillery rifle at A is aimed at an angle of 45o with the
horizontal, and its shell is just able to clear the mountain peak at the top of its
trajectory. Determine the magnitude u of the muzzle velocity, the height H of the
mountain above sea level, and the range R to the sea.
x
y
Horizontal:
Vertical:
SOLUTION
tvxxuvvaxooxoxx 45cos0
2
2
145sin gttvyygtugtvvga
yooyoyy
B
C
Point B (at apex) BByoy tutgugtvv 873.13045sin0
81.9
1
B
Bxoot
ututvxx11313.7
45cos08000
21 = smustB /18.39655.28
22 55.2881.92
155.2845cos18.3960
2
1 Byoo ygttvyy
myHmy BB 46006004000
Point C (sea level)
060014.280905.4
81.92
145cos18.3960600
2
1
2
2
2
CC
CC
yoo
tt
tt
gttvyy
06.2
18.59
Ct
mR
R
tvxxxoo
78.16578
18.5845cos18.3960
11. A projectile is launched with an initial speed of 200 m/s at an angle
of 60o with respect to the horizontal. Compute the range R as measured
up the incline.
x
y
s/m.sinv
s/mcosv
yo
xo
217360200
10060200
Horizontal:
tcosRtvxx xoo 100020 (1) at B
Vertical:
ay=gax=0
22 8192
12173020
2
1t.t.sinRgttvyy yoo (2) at B
(1) t=0.0094R
(2)
m.R
R..R..sinR
22967
009408192
100940217320
2
SOLUTION
12 . Determine the location h of the spot which the pitcher must throw
if the ball is to hit the catcher’s mitt. The ball is released with a speed of
40 m/s.
x
y
sinv
cosv
yo
xo
40
40
Horizontal:
cos
ttcostvxx xoo2
14020
Vertical:
ay=gax=0
22 8192
140081
2
1t.tsin.gttvyy yoo
586
6381
35516
02860
22612
574022614202005740202261
0812012261
1226120812
1819
2
1
2
14081
2
212
2
2222
.
.
.
.x
.
..x.xx.
xtan.tantan.
tansecsec.tan.cos
.cos
sin.
,
20 m
d1.638o
d=20 tan1.638=0.572 mh=2.8(1+0.572)=1.228 m
d
SOLUTION
13. A golfer strikes the ball at the top of a hill with an initial velocity of vo=12 m/s
and =60o with the horizontal. Determine the coordinates of the point the ball hits
the ground, its velocity at this moment and its total flight time. The equation of
curvature of the hill is given by y= 0.05x2. Dimensions are given in meters. Take
the gravitational attraction of the earth as g=9.81 m/s2.
vo
x
y=0.05x2
y
B
Determine the coordinates of the point the ball hits the ground, its velocity at this moment and its total
flight time.
vo=12 m/s
x60o
y=0.05x2
y
O
B
For B
For O
(t2 is the total flight time.)
Coordinates of B: xB=20.1 m , yB=20.24 m
Impact velocity: vx=6 m/s (cst), vy=(vo)y gt=10.39 9.81(3.35)=22.47 m/s
s/m.sinv
s/mcosv
yo
xo
39106012
66012
Horizontal: ttvxx xoo 6
Vertical:
ay=gax=0
2
2
90543910
2
1
t.t.y
gttvyy yoo
st
tttttt
tttxy
35.3
00105.339.1008.1905.439.10
605.0905.439.1005.0
2
1
222
222
s/m.vvv yx 262322