8 1. TOPIC D-1
b
d
y
D E
C
G
x
1
β
h
x
V
B
α
B
G
y
1
N
F
A
Figure 1.7.
1.3 SOLUTION
First, consider the motion of the object from point A to point B. Assume the object is a particle (because it is in
translational motion). Next, we show the forces acting on the object (particle): G
– weigh of the particle, N
– normal
reaction force, and F
– friction force. en, we write a dierential equation of motion of the particle on the distance
AB:
m
1
= ∑ F
xi
;
m
1
= G sin α – F.
e friction force:
F = μ
κ
N,
where N = G cos α.
→
9
en,
m
1
= G sin α – μ
κ
G cos α
or
1
= g sin α – μ
κ
g cos α.
Integrating this dierential equation twice, we will obtain:
1
= g(sin α – μ
κ
cos α )t + C
1
, (1.1)
x
1
=
g(sin α – μ
κ
cos α)
t
2
+ C
1
t + C
2
. (1.2)
To dene the integral constants, we will use the initial conditions of the problem. At t = 0, x
10
= 0, and
10
= V
A
= 0.
Using these initial conditions in Equations (1.1) and (1.2), we will obtain:
10
= C
1
and x
10
= C
2
.
en the integral constants:
C
1
= 0, C
2
= 0.
Substituting the values found for the integral constants in Equations (1.1) and (1.2):
1
= g(sin α – μ
κ
cos α)t, (1.3)
x
1
=
g(sin α
–
μ
κ
cos α)
t
2
. (1.4)
For the time instant τ when the object leaves the surface AB:
1
= V
B
and x
1
= l.
Substituting these values in Equations (1.3) and (1.4), we will obtain:
V
B
= g(sin α – μ
κ
cos α)τ, (1.5)
l =
g(sin α – μ
κ cos α)
τ
2
. (1.6)
Solving Equations (1.5) and (1.6) will yield:
V
B
=
2l
=
2 ∙ 4
= 8
m
⁄s.
τ
1
Now we will consider the motion of the object (particle) from point B to point C. e only force acting on the ob-
ject in this motion will be the gravity force (weigh) G
. e dierential equation of motion from point B to point C:
m = 0, (1.7)
mÿ = G. (1.8)
1.3 SOLUTION
2
2
2
10 1. TOPIC D-1
Integrating Equation (1.7):
= C
3
, (1.9)
x = C
3
t + C
4
. (1.10)
Using the initial conditions (at point B): at t = 0, x
0
= 0, and ẋ
0
= V
B
cos α. en from Equations (1.7) and (1.8)
for time t = 0:
0
= C
3
and x
0
= C
4
,
or C
3
= V
B
cos α; C
4
= 0.
Substituting the values of the integral constants in Equations (1.9) and (1.10):
= V
B
cos α, (1.11)
x = V
B
cos α ∙ t. (1.12)
Integrating Equation (1.8) twice:
= gt + C
5
, (1.13)
y =
gt
2
+ C
5
t + C
6
. (1.14)
2
e initial conditions at t = 0 (point B): y
0
= 0 and
0
= V
B
sin α. Using these initial conditions in Equations (1.13)
and (1.14), we will obtain:
0
= C
5
and y
0
= C
6
.
en, the integral constants: C
5
= V
B
sin α and C
6
= 0.
Substituting the values of the integral constants in Equations (1.13) and (1.14):
= gt + V
B
sin α,
y =
gt
2
+ V
B
sin α ∙ t.
2
Hence, the equations of motion of the object (particle):
x = V
B
cos α ∙ t, (1.15)
y =
gt
2
+ V
B
sin α ∙ t. (1.16)
2
e equation of the trajectory of the object will be found if we will eliminate the parameter t (time) in Equations
(1.15) and (1.16). Dening the parameter t from Equation (1.15) and substituting it in Equation (1.16), we will
obtain the equation of the parabola:
y =
gx
2
+ x tan α. (1.17)
2V
B
2
cos
2
α
11
At point C when the object landed:
y = h = 5 m and x = d.
Substituting these values for y and x in Equation (1.17):
5 =
9.81 d
2
+ d√3.
From here: d
1,2
= –2.82 ± 4.93, or d
1
= 2.11 m and d
2
= –7.75 m.
As the branch of the parabola is on the positive side of the abscissa, so we select d = 2.11 m.
e range b will be determined as:
b = d – ED = d –
h
= 2.11 –
5
= 0.77 m.
Using the equation of motion (1.15), we can nd the time of travel of the object from the point B to the point C (T):
2.11 = 8 ∙ 0.5 T.
From here:
T = 0.53 s.
To nd the velocity at the landing (at point C), we will use the equations of the velocity components (1.11) and
(1.13):
= V
B
cos α,
= gt + C
5
.
Substituting them in the equation:
v = √
2
+
2
at time t = T = 0.53:
V
C
= √(v
B
cos α)
2
+ (gT + V
B
sin α)
2
= √(8 ∙ 0.5)
2
+ (9.81 ∙ 0.53 + 8 ∙ 0.87)
2
= 12.8
m
.
s
1.3 SOLUTION
2 ∙ 8
2
∙ 0.5
2
tan 75° 3.73
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