Download 7. SSM REASONING According to Newton`s second

7. SSM REASONING According to Newton's second law of motion, the net force applied to the fist is equal to the mass of the fist multiplied by its acceleration. The data in the problem gives the final velocity of the fist and the time it takes to acquire that velocity. The average acceleration can be obtained directly from these data using the definition of average acceleration given in Equation 2.4. SOLUTION The magnitude of the average net force applied to the fist is, therefore, ⎛ Δv ⎞
⎛ 8.0 m/s – 0 m/s ⎞
∑ F = ma = m ⎜ ⎟ = ( 0.70 kg ) ⎜
⎟ = 37 N
0.15 s
⎝ Δt ⎠
⎝
⎠
11. REASONING Newton’s second law gives the acceleration as a = (ΣF)/m. Since we seek
only the horizontal acceleration, it is the x component of this equation that we will use;
ax = (ΣFx)/m. For completeness, however, the free-body diagram will include the vertical
forces also (the normal force FN and the weight W).
SOLUTION The free body diagram is
shown at the right, where
+y F1 = 59.0 N
F2 = 33.0 N
θ = 70.0°
When F1 is replaced by its x and y
components, we obtain the free body
diagram in the second drawing.
FN
F2
+x θ
W
F1
Choosing right to be the positive direction, we
have
ΣF
F cos θ − F2
ax = x = 1
m
m
ax =
( 59.0 N ) cos 70.0° − ( 33.0 N ) =
7.00 kg
+y
FN
−1.83 m/s 2
F2
The minus sign indicates that the horizontal
acceleration points to the left .
F1cos θ +x F1sin θ W
27. SSM REASONING AND SOLUTION a. According to Equation 4.4, the weight of an object of mass m on the surface of Mars would be given by W=
GM M m
RM2
where MM is the mass of Mars and RM is the radius of Mars. On the surface of Mars, the weight of the object can be given as W = mg (see Equation 4.5), so mg =
GM M m
RM2
or
g=
GM M
RM2
Substituting values, we have g=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(6.46 ×1023 kg)
= 3.75 m/s 2
6
2
(3.39 × 10 m)
b. According to Equation 4.5, W = mg = (65 kg)(3.75 m/s2) = 2.4 ×102 N
34. REASONING In each case the object is in equilibrium. According to Equation 4.9b, ΣFy = 0, the net force acting in the y (vertical) direction must be zero. The net force is composed of the weight of the object(s) and the normal force exerted on them. SOLUTION a. There are three vertical forces acting on the crate: an upward normal force +FN that the floor exerts, the weight –m1g of the crate, and the weight –m2g of the person standing on the crate. Since the weights act downward, they are assigned negative numbers. Setting the sum of these forces equal to zero gives FN + (−m1 g ) + (−m2 g ) = 0
Σ Fy
The magnitude of the normal force is FN = m1g + m2g = (35 kg + 65 kg)(9.80 m/s2) = 980 N
b. There are only two vertical forces acting on the person: an upward normal force +FN that the crate exerts and the weight –m2g of the person. Setting the sum of these forces equal to zero gives FN + (− m2 g ) = 0
ΣFy
The magnitude of the normal force is 2
FN = m2g = (65 kg)(9.80 m/s ) = 640 N 58. REASONING Since the mountain climber is at rest, she is in equilibrium and the net force
acting on her must be zero. Three forces comprise the net force, her weight, and the tension
forces from the left and right sides of the rope. We will resolve the forces into components
and set the sum of the x components and the sum of the y components separately equal to
zero. In so doing we will obtain two equations containing the unknown quantities, the
tension TL in the left side of the rope and the tension TR in the right side. These two
equations will be solved simultaneously to give values for the two unknowns.
SOLUTION Using W to denote the weight of the mountain climber and choosing right and
upward to be the positive directions, we have the following free-body diagram for the
climber:
For the x components of the forces we have
ΣFx = TR sin 80.0° − TL sin 65.0° = 0
+y
TL
For the y components of the forces we have
65.0º 80.0
ΣFy = TR cos80.0° + TL cos 65.0° − W = 0
TR +x W
Solving the first of these equations for TR, we find that
TR = TL
sin 65.0°
sin 80.0°
Substituting this result into the second equation gives
TL
sin 65.0°
cos80.0° + TL cos 65.0° − W = 0
sin 80.0°
or
TL = 1.717 W
Using this result in the expression for TR reveals that
TR = TL
sin 65.0°
sin 65.0°
= (1.717W )
= 1.580 W
sin 80.0°
sin 80.0°
Since the weight of the climber is W = 535 N, we find that
TL = 1.717 W = 1.717 ( 535 N ) = 919 N
TR = 1.580 W = 1.580 ( 535 N ) = 845 N
60. REASONING AND SOLUTION a. If the block is not to slide down the wall, then the vertical forces acting on the block must sum to zero. F cos 40.0° – mg + µsFN = 0
Additionally, the horizontal forces must sum to zero. F sin 40.0° – FN = 0
Eliminating FN gives F=
mg
= 79.0 N
cos40.0° + μ s sin 40.0°
b. The above analysis applies to the case where the block is starting to slide up the wall except, that the frictional force will be in the opposite direction. Hence, F=
mg
= 219 N
cos40.0° – μ s sin 40.0°
67.
REASONING AND SOLUTION
Newton's second law applied to object 1
(422 N) gives
Object 1
Object 2
N1
T = m1a1
T
T
Similarly, for object 2 (185 N)
W1
T – m2g = m2a2
W2
If the string is not to break or go
slack, both objects must have
accelerations of the same magnitude.
Then a1 = a and a2 = –a. The above equations become T = m1a T – m2g = – m2a (1) (2) a. Substituting Equation (1) into Equation (2) and solving for a yields a=
m2 g
= 2.99 m/s 2
m1 + m2
b. Using this value in Equation (1) gives T = m1 a = 129 N
81. REASONING The tension in each coupling bar is responsible for accelerating the objects behind it. The masses of the cars are m1, m2, and m3. We can use Newton’s second law to express the tension in each coupling bar, since friction is negligible: TA = ( m1 + m2 + m3 ) a
TB = ( m2 + m3 ) a
Coupling bar A
Coupling bar B
TC = m3a
Coupling bar C
In these expressions a = 0.12 m/s2 remains constant. Consequently, the tension in a given bar will change only if the total mass of the objects accelerated by that bar changes as a result of the luggage transfer. Using Δ (Greek capital delta) to denote a change in the usual fashion, we can express the changes in the above tensions as follows: ΔTA = ⎡⎣ Δ ( m1 + m2 + m3 ) ⎤⎦ a
ΔTB = ⎡⎣ Δ ( m2 + m3 ) ⎤⎦ a
Coupling bar A
Coupling bar B
ΔTC = ( Δm3 ) a
Coupling bar C
SOLUTION a. Moving luggage from car 2 to car 1 does not change the total mass m1 + m2 + m3, so Δ(m1 + m2 + m3) = 0 kg and ΔTA = 0 N . The transfer from car 2 to car 1 causes the total mass m2 + m3 to decrease by 39 kg, so Δ(m2 + m3) = –39 kg and (
)
ΔTB = ⎣⎡ Δ ( m2 + m3 ) ⎦⎤ a = ( –39 kg ) 0.12 m/s 2 = –4.7 N
The transfer from car 2 to car 1 does not change the mass m3, so Δm3 = 0 kg and ΔTC = 0 N . b. Moving luggage from car 2 to car 3 does not change the total mass m1 + m2 + m3, so Δ(m1 + m2 + m3) = 0 kg and ΔTA = 0 N . The transfer from car 2 to car 3 does not change the total mass m2 + m3, so Δ(m2 + m3) = 0 kg and ΔTB = 0 N . The transfer from car 2 to car 3 causes the mass m3 to increase by 39 kg, so Δm3 = +39 kg and (
)
ΔTC = ( Δm3 ) a = ( +39 kg ) 0.12 m/s 2 = +4.7 N