Newton`s Laws 2 body problems

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Newton’s Laws
and two body problems
• Our problems so far has been restricted to
one objects moving by a force. But what
happens if there are two objects connected
together in one way or another?
• We will find that the problem is solved in the
same general manner as when there is one
object - through the use of free-body
diagrams and Newton's laws.
Example problem 1:
• A 5.0-kg and a 10.0-kg box are touching each
other. A 45.0-N horizontal force is applied to the
5.0-kg box in order to accelerate both boxes
across the floor. Ignore friction and determine
the acceleration of the boxes and the force
acting between the boxes. (Contact Force)
The diagrams below show the freebody diagrams for the two objects.
Note that there are four forces on the
5.0-kg object at the rear.
• Using Fnet = m•a with the free-body diagram
for the 5.0-kg object will yield the Equation
below:
• 45.0 - Fcontact = 5.0•a
• Using Fnet = m•a with the free-body diagram
for the 10.0-kg object will yield the Equation
below:
• Fcontact = 10.0•a
• If the expression 10.0•a is substituted into
Equation 1 for Fcontact, then Equation 1
becomes reduced to a single equation with a
single unknown. The equation becomes
• 45.0 - 10.0•a = 5.0•a
• A couple of steps of algebra lead to an
acceleration value of 3.0 m/s2
• This value of a can be substituted back into
Equation 2 in order to determine the contact
force:
• Fcontact = 10.0•a = 10.0 •3.0
Fcontact = 30.0 N
Example Problem 2:
• A man enters an elevator holding two boxes one on top of the other. The top box has a mass
of 6.0 kg and the bottom box has a mass of 8.0
kg. The man sets the two boxes on a metric scale
sitting on the floor. When accelerating upward
from rest, the man observes that the scale reads
a value of 166 N; this is the upward force upon
the bottom box. Determine the acceleration of
the elevator (and boxes) and determine the
forces acting between the boxes.
The force of gravity is calculated in the
usual manner using 14.0 kg as the mass.
• Fgrav = m•g = 14.0 kg • 9.8 N/kg = 137.2 N
• Since there is a vertical acceleration, the
vertical forces will not be balanced; the Fgrav is
not equal to the Fnorm value.
• The net force is the vector sum of these two
forces. So
• Fnet = 166 N, up + 137.2 N, down = 28.8 N, up
• The acceleration can be calculated using
Newton's second law:
• a = Fnet /m = 28.8 N/14.0 kg = 2.0571 m/s2
• Now that acceleration has been determined,
using an individual object, either box, we can
determine the force acting between them.
• Fcontact - 58.8 N = (6.0 kg)•(2.0571 m/s2)
Pulley Problems (omg)
• 1. Consider the two-body situation below. A
100.0-gram hanging mass (m2) is attached to
a 325.0-gram mass (m1) at rest on the table.
The coefficient of friction between the 325.0gram mass and the table is 0.215. Determine
the acceleration of the system and the tension
in the string.
• a = 0.695 m/s2 and Ftens = 0.911 N
• The solution here will use the approach of a
free-body diagram and Newton's second law
analysis of each individual mass.
•
• For mass m1:
Fgrav = m1•g = (0.3250 kg)•(9.8 N/kg) = 3.185 N
Fnorm = Fgrav = 3.185 N
Ffrict = µ•Fnorm = (.215)•(3.185 N) = 0.68478 N
• Applying Newton's second law to m1:
Ftens - 0.68478 = 0.3250•a
• For mass m2:
Fgrav = m2•g = (0.1000 kg)•(9.8 N/kg) = 0.9800 N
• Applying Newton's second law to m2:
0.9800 - Ftens = 0.1000•a
• Combining the two equations leads to the
answers:
a = 0.69464 m/s2 = ~0.695 m/s2
Ftens = 0.91054 N = ~0.911 N
Example 2 Pulley
• Consider the two-body situation at the right. A
3.50x103-kg crate (m1) rests on an inclined plane
and is connected by a cable to a 1.00x103-kg mass
(m2). This second mass (m2) is suspended over a
pulley. The incline angle is 30.0° and the surface
has a coefficient of friction of 0.210. Determine the
acceleration of the system and the tension in the
cable.
Because the parallel component of gravity on m1
exceeds the sum of the force of gravity on m2 and the
force of friction, the mass on the inclined plane (m1)
will accelerate down it and the hanging mass (m2) will
accelerate upward.
• For mass m1:
Fgrav = m1•g = (3500 kg)•(9.8 N/kg) = 34300 N
Fparallel = m1•g•sine(θ) = 34300 N • sine(30°) =
17150 N
Fperpendicular = m1•g•cosine(θ) = 34300 N •
cosine(30°) = 29704.67 N
Fnorm = Fperpendicular = 29704.67 N
Ffrict = µ•Fnorm = (.210)•(29704.67 N) = 6237.98 N
• Applying Newton's second law to m1:
17150 N - 6237.98 N - Ftens = (3500 kg)•a
• For mass m2:
Fgrav = m2•g = (1000 kg)•(9.8 N/kg) = 9800 N
• Applying Newton's second law to m2:
Ftens - 9800 N - = (1000 kg)•a
• Combining the two equations leads to the
answers:
a = 0.24712 m/s2 = ~0.247 m/s2
Ftens = 10047 N = ~1.00x104 N

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