## Connected Objects

Example:Two blocks of masses m_{1}and m_{2}are placed in contact with each other on a smooth, horizontal plane as shown here. A constant horizontal forceFis applied to m_{1}. What is the acceleration of each mass?In one sense, we can (almost) solve this example intuitively -- in our head. A force F is applied to an object whose mass is m = m

_{1}+ m_{2}. So its acceleration must bea = F / m or

a = F / (m _{1}+ m_{2})That's the right answer! But is there nothing more to this question? Simple questions -- the intuitively obvious kind -- make wonderful templates or examples for more difficult problems.

Look at all the forces on m

_{1}. Make a good free-body diagram of the forces acting on m_{1}.Of course the external force

Fpushes to the right on the mass m_{1}. Gravity pulls down with forcew_{1}= m_{1}gand the plane responds with a normal forcen_{1}. But the other mass -- m_{2}-- exerts a force on mass m_{1}. This force is labeledP'and points to the left. We can apply Newton's Second Law to the y-component forces and find that n_{1}= w_{1}1. But now there is an additional and unknown force in the x-component of Newton's Second Law,F _{1,net}= F - P' = m_{1}aWe need

more informationso we turn to the other mass, m_{2}The first mass m1 exerts a force

Pon this mass, m_{2}. Applying the y-component ofF= ma, we readily findn _{2}= w_{2}For the x-components, the

onlyforce acting on m_{2}is P soF _{2,net}= P = m_{2}aHowever, from Newton's Third Law,

F_{12}= -F_{21}, we know that P = P' soF - P' = m _{1}aF - m

_{2}a = m_{1}aF = m

_{1}a + m_{2}aF = ( m

_{1}+ m_{2}) aa = F / ( m

_{1}+ m_{2})That is the same answer we found so quickly earlier but this may provide a pattern to use for more complex situations.

Example:This particular example is stated in terms of weighing afishin an accelerating elevator. It is also fun to think of weighingyourselfin an accelerating elevator. When does an elevatoraccelerate upwards? When does an elevatoraccelerate downwards?The forces acting on the fish are shown in the free-body diagram. T is the tension supplied by the scale. This is the value the scale reads. We may call it the

apparent weightof the fish. The net force on the fish isF _{net}= T - wor

F _{net}= T - m gThe net force is (

always!) equal to the mass times the acceleration. This fish is moving along with the elevator. In this diagram we have taken the acceleration to beupso it is positive.F _{net}= T - m g = m aT = m g+ m a

T = m (g + a)

While the elevator accelerates

upward, theapparent weightof the fish isgreaterthan its true weight, mg.What happens as the elevator accelerates

downward?The forces on the fish are again shown in the free-body diagram,

F _{net}= T - wor

F _{net}= T - mgThe net force is (

always!) equal to the mass times the acceleration. This fish is moving along with the elevator. Now the acceleration to beupso it isnegative.F _{net}= T - m g = m ( - a )T = m g - m a

T = m (g - a)

While the elevator accelerates

downward, theapparent weightof the fish islessthan its true weight, mg.Try this

yourselfon an elevator -- not weighing a fish, but paying attention to your ownapparent weight!

(c) Doug Davis, 2005; all rights reserved