PHYSICS 1: Solutions to: Classical Mechanics

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PHYSICS 1Solutions toClassical Mechanics

• Ronald L. Westra PhD
• Dept. Mathematics
• Maastricht University

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1. STATICS
Q2 At what angle a will the object start to
slide down?
INCLINED PLANE
angle a
Answer When tan a f
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1. STATICS
Q3 How large must F be such that the system is
in equilibrium?
F ?
B
armlength OB 1.5 m
A
armlength OA 1 m
FR (reaction force)
Fg (weight) m.g 100 N
angle a 30o
O
Here Fg 100 N, armlength OA 1 m ,
armlength OB 1.5 m, a 30 o
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1. STATICS
Q3 How large must F be such that the system is
in equilibrium?
F ?
F2 F.cos a
B
armlength OB 1.5 m
A
armlength OA 1 m
F1 F sin a
FR (reaction force)
FR2
angle a 30o
Fg (weight) m.g 100 N
O
FR1
STEP 1 Select reference frame O as origin,
horizontalvertical coordinates STEP 2
decompose the forces STEP 3 static forces
X-axis FR1 (- F sin a) 0
1 static forces Y-axis FR2 F cos a -
Fg 0 2 STEP 4 static
torque relative to O F.OB FgOA.cos a 0
3 Three linear equations with three variables
(F, FR1, FR2) solvable!
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Q3 How large must F be such that the system is
in equilibrium?
FR1 (- F sin a) 0
1 FR2 F cos a - Fg 0
2 F.OB FgOA.cos a 0 3 These
solve to FR1 Fg sin(2a)/3 ( 34.2 N) FR2
Fg (1 2cos²(a)/3) ( 35.3 N) F
2(Fgcos a)/3 ( 57.7 N) Notice that FR2/FR1
(2 - cos 2a)/sin 2a gt tan a, so the reaction
force FR does not point in the direction of arm
OB but above it!
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2. KINEMATICS
Q4 Describe the motion of this object.
rain 0.01 kg/sec
FN (Normal force) 10 N
v 10 m/s
Ff (friction force) 0 N
Fg (weight) 10 N
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Q4 Describe the motion of this object.
There is only change in the horizontal
coordinate as the mass slowly increases as
m(t) 1 0.01.t, (NB g 10 m/s²) and there
is no net force. Now apply Newtons second law
d(m(t)v(t))/dt 0 ? m(t)v(t) constant
m(0)v(0) 10 v(t) 10/(1 0.01.t)
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2. KINEMATICS
Q5 Describe the motion of this object.
Ff (friction force) -0.1v
Fg (weight) 0.1 N
A rain drop falls freely under the force of
gravity. The force of air friction Ff is
directed opposite to the velocity of the droplet,
and has a size that is a constant (0.1 kg/sec)
times the magnitude of this velocity Ff
-0.1 v
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Q5 Describe the motion of this object.
There is only movement in the vertical direction
choose there your axis. The force that act
are gravity (0.1N) and air friction (-0.1.v), so
the total force points downward (hence is
negative) and equals Ftot - 0.1 0.1.v.
Let g 10 m/s² and the initial velocity at t0
be 0 m/s. Now apply Newtons second
law d(m(t)v(t))/dt Ftot - 0.1 0.1.v ?
use some calculus v(t) -100
100exp(-0.1.t) So, after some time the rain
drop falls freely under the force of gravity
with a constant velocity of 100 m/s
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Q5 Describe the motion of this object.
v(t) -100 100exp(-0.1.t)
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2. KINEMATICS
Q6 Describe the motion of this object.
target
100 m
A ballista launches a rock with 50 m/s in a
uniform gravitational field with g -10 m/s2.
What is the safest angle a with the horizon
in order to hit a target at a distance of 100
m? And what information is (not) required?
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Q6 Describe the motion of this object.
Fg (weight) 10 N g 10 m/s2
Here we have two independent movements Horizont
al Ftot 0, so d(mv1(t))/dt 0 ? v1(t)
constant v1(0) v cos a Vertical Ftot -10,
so d(mv2(t))/dt -10 ? v2(t) v2(0) 10t (v
sin a) 10t Therefore the vertical position z
is z(t) (v sin a)t 5t² . The bullet hits
the ground if z0, so when (v sin a)t 5t² 0,
i.e. at t0 (trivial, the start) and tmax (v
sin a)/5. The vertical position is then x(tmax)
(v cos a)tmax v² sin(2a)/10 250 sin(2a).
This must be equal to the position of the goal
100 m, so x(tmax) 250 sin(2a) 100 ? sin(2a)
0.4. This means that 2a 23.5782 or 2a 180 -
23.5782 156.4218. So a 11.7891 or a
78.2109. The smaller angle is better, for it
remains less long in the air and therefore is
more accurate.