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Some Applications of Newton’s Laws.

In this chapter, we get some practice applying Newton’s laws to various physical problems. We

do not introduce any new laws of physics.

Solving Fnet = ma problems with multiple bodies

Problem: A mass m1 is pulled up a frictionless incline by a string over a pulley and a hanging

mass m2.

We know m1, m2, and the angle .

We seek :

T = tension in the cord,

a = acceleration of the mass,

N = normal force on m1

Step 1

Draw a free-body diagram for each moving object.

Label the force arrows with

the magnitudes of the forces.

Notice that T is the same for

both objects (by NIII).

Use m1, m2, in the diagrams,

not m.

Step 2

For each object, choose xy axes so that the acceleration vector a is in the (+) direction.

Notice that we can choose

different axes for the different

objects. And we can tilt the

axes if necessary.

m1

m2

no friction

T

N

m1g

T

m2g

T

N

m1g

T

m2g

+x

+y

a

+y

a

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Step 3 For each object, write the equations

x x y y

F m a , F m a

m1 :

11

1

x-eqn (1) T m gsin m a

y-eqn (2) N m gcos 0

m2 :

22

(3) m g T m a

Notice that

aa

is the same for both m1 and m2 since they are connected by a string that

doesn't stretch.

Now we have a messy algebra problem with 3 equations in 3 unknowns:

11

1

22

(1) T m gsin m a

(2) N m gcos 0

(3) m g T m a

The unknowns are T , N , and a.

We can solve for N right away. Eqn (2)

1

N m gcos

Now we have 2 equations [(1) & (3)] in 2 unknowns ( a & T ).

Solve (1) for T and plug into (3) to get an equation without an T :

(1)

1 1 1

T m gsin m a m (gsin a)

(3)

2 1 2

m g [m (gsin a)] m a

T

N

m1g

T

m2g

x

y

+y

a

m1g

m1g cos

m1g sin

2

1

a

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Now solve this last equation for a:

Finally, if you have any strength left, we can solve for tension T by plugging our expression for

acceleration a back into either (1) or (3) and solving for T. From (3), we have

2 2 2

T m g m a m (g a)

.

Plugging in our big expression for a, we get

21

22

12

m g m gsin

T m (g a) m g (m m )

We can simplify:

1 2 2 1 1 2 2 1

22

1 2 1 2 1 2

1 1 2 1

2

1 2 1 2

g(m m ) m g m gsin m g m g m g m gsin

T m m

(m m ) (m m ) (m m )

m g m gsin m m g

T m T (1 sin )

(m m ) (m m )

Do these expressions make sense? Let’s check some limits.

If m1 = 0, then m2 should be freely falling with a = g and the tension T should be zero. Check

that this is so.

If m2 = 0, then m1 should slide down the incline with acceleration a = –g sin (since it would be

accelerating in the negative direction). Check that this is so.

2 1 1 2 1 2

21

12

m g m gsin m a m a a(m m )

m g m gsin

a(m m )

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