Download aguilar (fa6754) – hk5 – opyrchal – (11106)

aguilar (fa6754) – hk5 – opyrchal – (11106)
This print-out should have 6 questions.
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before answering.
001 10.0 points
Two men decide to use their cars to pull a
truck stuck in mud. They attach ropes and
one pulls with a force of 758 N at an angle
of 31◦ with respect to the direction in which
the truck is headed, while the other car pulls
with a force of 1203 N at an angle of 22◦ with
respect to the same direction.
75
8N
1
002 10.0 points
Two forces are the only forces acting on a
4.3 kg object which moves with an acceleration of 3.9 m/s2 in the positive y direction.
One of the forces acts in the positive x direction and has a magnitude of 10 N.
What is the magnitude of the other force
f2 ?
Correct answer: 19.5252 N.
Explanation:
Basic Concepts:
◦
31
X
F = ma
22 ◦
120
3N
Solution:
What is the net forward force exerted on
the truck in the direction it is headed?
Correct answer: 1765.13 N.
f2
f1
f~2 is the hypotenuse of a right triangle, so
Explanation:
f~2 =
F1
F2
θ1
θ2
= 758 N ,
= 1203 N ,
= 31◦ , and
= 22◦ .
For the first vehicle, the forward component
is
F1f = F1 cos θ1
= (758 N) cos 31◦
= 649.733 N .
003 10.0 points
A 7.6 kg object hangs at one end of a rope that
is attached to a support on a railroad boxcar.
When the car accelerates to the right, the
rope makes an angle of 29◦ with the vertical
The acceleration of gravity is 9.8 m/s2 .
Similarly, for the second vehicle,
F2f = F2 cos θ2
= (1203 N) cos 22◦
= 1115.4 N .
Thus the net forward force on the truck is
Ff = F1f + F2f
= 649.733 N + 1115.4 N
= 1765.13 N .
q
f 2 + f1 2
29 ◦
Let :
f
a
7.6 kg
Find the acceleration of the car. (Hint:
~aobject = ~acar )
Correct answer: 5.43223 m/s2 .
Explanation:
aguilar (fa6754) – hk5 – opyrchal – (11106)
Given :
Explanation:
m = 7.6 kg ,
θ = 29◦ , and
Let :
g = 9.8 m/s2 .
T cos θ θ
2
T
a = 3.6 m/s2 ,
θ = 30◦ , and
g = 9.81 m/s2 .
Consider the free body diagram for the
block
a
T sin θ
mg
Vertically
X
Fy = T cos θ − m g = 0
T cos θ = m g .
(1)
Horizontally,
X
Fx = T sin θ = m a .
m
in
gs
N
θ
os
µN g c
m
=
θ
mg
Consider the forces parallel to the ramp
(2)
Fk = Fg,x − m ax
= m g sin θ − m ax
Dividing Eqs 1 and 2, we have
T sin θ
a
=
T cos θ
g
a
tan θ =
g
a = g tan θ
= 9.8 m/s2 tan 29◦
= 5.43223 m/s2 .
004 10.0 points
A block accelerates at 3.6 m/s2 down a plane
inclined at an angle 30.0◦ .
and the forces perpendicular to the ramp
Fn = Fg,y = m g cos θ .
The coefficient of friction is
m g sin θ − m a
Fk
=
Fn
m g cos θ
g sin θ − a
=
g cos θ
9.81 m/s2 sin 30◦ − 3.6 m/s2
=
9.81 m/s2 cos 30◦
µk =
= 0.153607 .
2
m
3. 6
/s
005 10.0 points
The board sandwiched between two other
boards in the figure weighs 96.0 N.
m
µk
30◦
96 N
Find µk between the block and the inclined plane. The acceleration of gravity is
9.81 m/s2 .
Correct answer: 0.153607.
If the coefficient of friction between the
boards is 0.518, what must be the magnitude
aguilar (fa6754) – hk5 – opyrchal – (11106)
3
of the horizontal forces acting on both sides
of the center board to keep it from slipping
downward?
A 27 N force is required to push the block
up the incline with constant velocity.
What is the weight of the block?
Correct answer: 92.6641 N.
Correct answer: 35.2515 N.
Explanation:
Explanation:
Fs
Fs
Let : : F = 27 N ,
θ = 26◦ ,
α = 11◦ ,
µ = 0.3 , and
W = mg.
96 N
Note: Figure is not drawn to scale.
Basic Concepts: The horizontal forces
are the normal forces.
Consider the free body diagram for the
block
Fs,max = µs Fn
F
µN
Fy,net = 2 Fs,max − Fg = 2 µs Fn − Fg = 0 .
N
Given:
mg
Fg = 96.0 N
µs = 0.518 .
Perpendicular to the plane, the force component F sin α and the weight component
W cos θ act into the plane and the normal
force N acts out of the plane:
Solution:
2 µs Fn = Fg
Fg
Fn =
2 µs
(96 N)
=
2 (0.518)
N − F sin α − W cos θ = 0
=⇒ N = F sin α + W cos θ
= 92.6641 N .
006 10.0 points
A block is placed on a inclined plane.
The acceleration of gravity is 9.8 m/s2 .
27 N
11◦
Parallel to the plane, the force component
F cos α acts up the plane and the weight
component W sin θ and friction act down the
plane:
F cos α − W sin θ − µ N = 0
F cos α − W sin θ − µ (F sin α + W cos θ) = 0
F cos α − µ F sin α = W sin θ + µ W cos θ
F cos α − µ F sin α
sin θ + µ cos θ
(27 N) cos 11◦ − (0.3) (27 N) sin 11◦
=
sin 26◦ + (0.3) cos 26◦
= 35.2515 N .
W=
W
µ=
0. 3
26◦