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Mg 216 FORCES AND NEWTON'S LAWS OF MOTION In these applications we have taken upward as the positive direction, so that Robin’s
acceleration is a, while the chandelier’s acceleration is –a. Solving the RobinHood
equation for T gives
T = mg + ma
Substituting this expression for T into the Chandelier equation gives mg + ma – Mg = – Ma M –m
a=
g
M +m or a. Robin’s acceleration is (195 kg ) – ( 77.0 kg ) M –m
2
2
a= 9.80 m/s = 4.25 m/s
g = M +m
195 kg ) + ( 77.0 kg ) ( ( ) b. Substituting the value of a into the expression for T gives ( ) T = mg + ma = ( 77.0 kg ) 9.80 m/s 2 + 4.25 m/s2 = 1080 N
____________________________________________________________________________________________ 84. REASONING Newton’s second law, Equation 4.2a, can be used to find the tension in the
coupling between the cars, since the mass and acceleration are known. The tension in the
coupling between the 30th and 31st cars is responsible for providing the acceleration for the
20 cars from the 31st to the 50th car. The tension in the coupling between the 49th and 50th
cars is responsible only for pulling one car, the 50th.
SOLUTION
a. The tension T between the 30th and 31st cars is
Tx = (Mass of 20 cars)a x ( (4.2a) )( ) = ( 20 cars ) 6.8 × 103 kg/car 8.0 × 10−2 m / s 2 = 1.1 × 104 N
b. The tension T between the 49th and 50th cars is
Tx = (Mass of 1 car)a x ( (4.2a) )( ) = (1 car ) 6.8 × 103 kg/car 8.0 × 10−2 m / s 2 = 5.4 × 102 N
____________________________________________________________________________________________ Chapter 4 Problems 85. REASONING The box comes to a halt because the kinetic
FN
frictional force and the component of its weight parallel to the
incline oppose the motion and cause the box to slow down. The
distance that the box travels up the incline can be can be found
mg sinθ
2
by solving Equation 2.9 ( v 2 = v0 + 2ax ) for x. Before we use
this approach, however, we must first determine the acceleration
fk
of the box as it travels along the incline. 217 mg cos θ SOLUTION The figure above shows the freebody diagram for the box. It shows the
resolved components of the forces that act on the box. If we take the direction up the incline
as the positive x direction, then, Newton's second law gives
∑ Fx = – mg sin θ – f k = max – mg sin θ – µk FN = max or where we have used Equation 4.8, f k = µk FN . In the y direction we have
∑ Fy = FN – mg cos θ = 0 or FN = mg cos θ since there is no acceleration in the y direction. Therefore, the equation for the motion in
the x direction becomes
– mg sin θ – µ k mg cos θ = ma x or ax = – g (sin θ + µ k cos θ ) According to Equation 2.9, with this value for the acceleration and the fact that v = 0 m/s,
the distance that the box slides up the incline is x=– 2
v0 2a = 2
v0 2 g (sin θ + µk cos θ ) = (1.50 m/s)2
= 0.265 m
2(9.80 m/s 2 )[sin 15.0° + (0.180)cos 15.0°] ____________________________________________________________________________________________ 86. REASONING Since we assume that there is no frictional
force resisting the airplane’s motion, the only horizontal
force acting on the airplane arises because of the tension
(magnitude = T) in the cable. The airplane (mass = M)
undergoes a horizontal acceleration caused by the
horizontal component Tx = T cos θ of the tension force,
where θ is the angle that the cable makes with the
horizontal. From Newton’s second law, the acceleration of
the airplane is ax = ΣFx
M = T cos θ
M +y FN
fsMAX Tx
θ T +x
Ty mg
(1)
Freebody diagram
of the man 218 FORCES AND NEWTON'S LAWS OF MOTION The maximum tension in the cable is limited by the condition that the man’s feet must not
slip. When the man pulls as hard as possible without slipping, the horizontal component of
the tension acting on him matches the maximum static frictional force: Tx = T cos θ = fsMAX
(see the freebody diagram of the man). The maximum static frictional force itself is given
by fsMAX = µs FN (Equation 4.7). Together, these two relations yield
T cos θ = µs FN (2) To evaluate the magnitude FN of the normal force that acts on the man, we must consider
Newton’s third law. This law indicates that when the man (mass = m) pulls up on the cable,
the cable pulls down on him (see the freebody diagram of the man). This additional
downward force increases the upward normal force FN the runway exerts on him. Applying
Newton’s second law to the vertical direction in this diagram, with zero acceleration, we see
that ΣFy = FN − Ty − mg = FN − T sin θ − mg = 0 . Solving for FN yields FN = T sin θ + mg (3) Substituting Equation (3) into Equation (2) yields an expression in which the tension T is the
only unknown quantity: T cos θ = µs (T sin θ + mg ) (4) We now solve Equation (4) for the tension in the cable:
T cos θ = µsT sin θ + µs mg or T cos θ − µsT sin θ = µs mg T= or T ( cos θ − µs sin θ ) = µs mg µs mg
cos θ − µs sin θ (5) Equation (5) may be substituted into Eq...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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