Version 001 – HW#3 - Forces – arts – (00223)
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Hanging Weight 06
001 (part 1 of 2) 10.0 points
Consider the 624 N weight held by two cables
shown below. The left-hand cable had tension T2 and makes an angle of 48 ◦ with the
ceiling. The right-hand cable had tension T1
and makes an angle of 37◦ with the ceiling.
1
we have
x
Fnet
= F1x − F2x = 0
= F1 cos θ1 − F2 cos θ2 = 0
(1)
y
y
y
Fnet = F1 + F2 − Wg = 0
= F1 sin θ1 + F2 sin θ2 − Wg = 0 (2)
Solution: Using Eq. 1, we have
F2 = F1
cos θ1
.
cos θ2
Substituting F2 from Eq. 2 into Eq. 1, we have
48 ◦
◦
37
T2
T1
624 N
a) What is the tension in the cable labeled
T1 slanted at an angle of 37◦ ?
Correct answer: 419.132 N.
Explanation:
Observe the free-body diagram below.
002 (part 2 of 2) 10.0 points
a) What is the tension in the cable labeled T2
slanted at an angle of 48◦ ?
Correct answer: 500.252 N.
F2
θ2
F1 sin θ1 + F2 sin θ2 = Wg
cos θ1
F1 sin θ1 + F1
sin θ2 = Wg
cos θ2
F1 sin θ1 + F1 cos θ1 tan θ2 = Wg
Wg
F1 =
sin θ1 + cos θ1 tan θ2
624 N
F1 =
◦
sin 37 + cos 37◦ tan 48◦
= 419.132 N
F1
θ1
Explanation:
Using Eq. 1, we have
F2 = F1
Wg
Note: The sum of the x- and
y-components of F1 , F2 , and
Wg are equal to zero.
Given : Wg = 624 N ,
θ1 = 37◦ , and
θ2 = 48◦ .
Basic Concept: Vertically and Horizontally,
cos θ1
cos θ2
cos 37◦
= (419.132 N)
cos 48◦
= 500.252 N .
Static Equilibrium 05
003 (part 1 of 2) 10.0 points
The knot at the junction is in equilibrium
under the influence of four forces acting on
it. The F force acts from above on the left at
an angle of α with the horizontal. The 6.6 N
force acts from above on the right at an angle
of 45 ◦ with the horizontal. The 5.9 N force
Version 001 – HW#3 - Forces – arts – (00223)
acts from below on the right at an angle of
43◦ with the horizontal. The 6.3 N force acts
from below on the left at an angle of 49◦ with
the horizontal.
6
N
α
F
6.
45 ◦
knot
N
3
N
43
9
5.
6.
◦
49 ◦
Note: F, α are not to scale.
What is the magnitude of the force F ?
Correct answer: 6.35728 N.
Explanation:
Note: Standard angular measurements are
from the positive x-axis in a counter-clockwise
direction.
Given : F1
α1
F2
α2
F3
α3
F4
α4
= 6.6 N ,
= 180◦ − 45◦ = 135◦ ,
= 5.9 N ,
= 180◦ + 43◦ = 223◦ ,
= 6.3 N ,
= 360◦ − 49◦ = 311◦ ,
= F , and
= α.
Observe the free-body diagram below
where the vectors are decomposed into components along the x- and y-axes.
6
N
9
N
3
3
N
5.
6.
F1y + F2y + F3y + F4y = 0
F1 sin α1 + F2 sin α2
+F3 sin α3 + F4y = 0
40. 3
49 ◦
Note: The vectors along the xand y-coordinates add to zero.
Basic Concepts:
X
Fx = 0
(1)
(2)
Solution: Using Eqs. 1 and 2, we have
F4x = −F1 cos α1 − F2 cos α2
− F3 cos α3
(1)
◦
◦
= −(6.6 N) cos 135 − (5.9 N) cos 223
− (6.3 N) cos 311◦
= 4.84872 N , and
F4y = −F1 sin α1 − F2 sin α2
− F3 sin α3
(2)
◦
◦
= −(6.6 N) sin 135 − (5.9 N) sin 223
− (6.3 N) sin 311◦
= 4.11156 N , so
q
F4 = (F4x )2 + (F4y )2
q
= (4.84872 N)2 + (4.11156 N)2
= 6.35728 N .
004 (part 2 of 2) 10.0 points
What is the angle α of the force F as shown
in the figure above?
Correct answer: 40.2969◦ .
6.
◦
43
6.
45 ◦
6N
F1x + F2x + F3x + F4x = 0
F1 cos α1 + F2 cos α2
+F3 cos α3 + F4x = 0
X
Fy = 0
2
Explanation:
F4y
α4 = arctan
Fx
4
4.11156 N
= arctan
4.84872 N
◦
= 135 , from the positive x−axis
α = 40.2969◦ .
Accelerated System 01
Version 001 – HW#3 - Forces – arts – (00223)
005 (part 1 of 2) 10.0 points
A block of mass 5.42 kg lies on a frictionless
horizontal surface. The block is connected by
a cord passing over a pulley to another block
of mass 7.47 kg which hangs in the air, as
shown. Assume the cord to be light (massless
and weightless) and unstretchable and the
pulley to have no friction and no rotational
inertia.
5.42 kg
3
own weight W2 = m2 g (downward). Consequently,
m2 a = F2net↓ = m2 g − T .
Adding,
(m1 + m2 ) a = m2 g
m2
g
a=
m1 + m2
7.47 kg
=
(9.8 m/s2 )
5.42 kg + 7.47 kg
= 5.67929 m/s2 .
7.47 kg
Calculate the acceleration of the first block.
The acceleration of gravity is 9.8 m/s2 .
Correct answer: 30.7817 N.
Correct answer: 5.67929 m/s2 .
Explanation:
Explanation:
a
= 30.7817 N .
T
T
m1
N
T = m1 a = (5.42 kg) (5.67929 m/s2 )
m1 = 5.42 kg and
m2 = 7.47 kg .
m2
m1 g
a
m2 g
Since the cord is unstretchable, the first
block accelerates to the right at exactly the
same rate a as the second (hanging) block accelerates downward. Also, the cord’s tension
pulls the first block to the right with exactly
the same tension T as it pulls the second block
upward.
The only horizontal force acting on the first
block is the cord’s tension T ,so by Newton’s
Second Law
m1 a = F1net→ = T .
The second block feels two vertical forces:
The cord’s tension T (upward) and the block’s
Pulling Two Blocks 02
007 10.0 points
Two blocks connected by a string are pulled
across a rough horizontal surface by a force
applied to one of the blocks, as shown.
The acceleration of gravity is 9.8 m/s2 .
F
Let :
006 (part 2 of 2) 10.0 points
Calculate the tension in the cord.
3 kg
T
◦
59
9 kg
µ = 0.2
If each block has an acceleration of 2.2 m/s2
to the right, what is the magnitude of the
applied force?
Correct answer: 72.7197 N.
Explanation:
Version 001 – HW#3 - Forces – arts – (00223)
Given : M1 = 3 kg ,
M2 = 9 kg ,
θ = 59◦ ,
a = 2.2 m/s2 ,
µ = 0.2 .
For the mass M1 , N1 = W1 ,
4
Correct answer: 0.47907.
Explanation:
Consider the free body diagram:
a
and
fk
θ
so
Fnet = M1 a = T − µ M1 g .
(1)
For the mass M2 ,
N2 + F sin θ = W2 = M2 g
N2 = M2 g − F sin θ , so (2)
Fnet = M2 a
= F cos θ − T
− µ (M2 g − F sin θ) . (3)
First find the normal force acting on the
box using
X
Fy = 0.
So,
N − m g cos θ = 0.
Now apply Newton’s 2nd Law to the motion
parallel to the ramp.
X
F = m g sin θ − fk = m a.
Adding Eqs. 1 and 3, we have
And
(M1 + M2 ) a = F cos θ − µ M1 g − µ M2 g
+ µ F sin θ .
(M1 + M2 ) (a + µ g)
cos θ + µ sin θ
3 kg + 9 kg
=
cos 59◦ + (0.2) sin 59◦
× 2.2 m/s2 + (0.2) 9.8 m/s2
fk = µ N = µ m g cos θ.
Then,
m g sin θ − µ m g cos θ = m a
F =
= 72.7197 N .
Using Eq. 2, we have
N2 = M2 g − F sin θ
= (9 kg) (9.8 m/s2 ) − (72.7197 N) sin(59◦ )
= 25.867 N .
Sliding Down a Ramp
008 10.0 points
A box slides down a 33◦ ramp with an acceleration of 1.4 m/s2 .
The acceleration of gravity is 9.8 m/s2 .
Determine the coefficient of kinetic friction
between the box and the ramp.
µ m g cos θ = m g sin θ − ma
so
g sin θ − a
g cos θ
(9.8 m/s2 ) sin 33◦ − 1.4 m/s2
=
(9.8 m/s2 ) cos 33◦
= 0.47907
µ=
Acceleration with Friction 01
009 10.0 points
There is friction between the block and the
table.
The suspended 2 kg mass on the left is
moving up, the 2 kg mass slides to the right
on the table, and the suspended mass 4 kg on
the right is moving down.
The acceleration of gravity is 9.8 m/s2 .
Version 001 – HW#3 - Forces – arts – (00223)
2 kg
For the mass m1 , T1 acts up and the weight
m1 g acts down, with the acceleration a directed upward, so
µ = 0.11
Fnet1 = m1 a = T1 − m1 g .
2 kg
4 kg
What is the magnitude of the acceleration
of the system?
Explanation:
a
For the mass on the table, a is directed to
the right, T2 acts to the right, T1 acts to the
left, and the motion is to the right so that the
frictional force µ m2 g acts to the left and
Fnet2 = m2 a = T2 − T1 − µ m2 g .
Fnet3 = m3 a = m3 g − T2 .
m2
m3 − µ m2 − m1
g
m1 + m2 + m3
4 kg − (0.11) (2 kg) − 2 kg
=
2 kg + 2 kg + 4 kg
× (9.8 m/s2 )
m3
m1 = 2 kg ,
m2 = 2 kg ,
m3 = 4 kg ,
µ = 0.11 .
a=
= 2.1805 m/s2 .
and
Basic Concepts: The acceleration a of
each mass is the same, but the tensions in the
two strings will be different.
Fnet = m a 6= 0
Solution: Let T1 be the tension in the left
string and T2 be the tension in the right string.
Consider the free body diagrams for each
mass
N
µN
T2
a
T1
m1 g
(3)
(m1 + m2 + m3 ) a = m3 g − µ m2 g − m1 g
m1
T1
(2)
Adding these equations yields
µ
a
(1)
For the mass m3 , T2 acts up and the weight
m3 g acts down, with the acceleration a directed downward, so
Correct answer: 2.1805 m/s2 .
Let :
5
a
T2
m2 g
m3 g