CK-12-Physics - Intermediate

5.2. Inclined Planes http://www.ck12.org

They−component ofMg:Mgcosθ

Newton’s Second Law equation for the massm:

We assume thatmgmoves down and thereforemg>T. (This is an arbitrary choice.)

∑
m

F=mg−T=ma;→∑

m

F= (30kg)(10m/s^2 )−T= 30 a

Newton’s Second Law equation forM

Sincemmoves down,Mmoves up the incline. ThereforeT>Mg.

∑
M

F=T−Mgsinθ=Ma→∑

M

F=T−( 40 )( 9. 8 )sin 25◦= 40 a

Solving the two equations above simultaneously for the acceleration gives:

( 30 )( 9. 8 )−T+T−( 40 )( 9. 8 )sin 25◦= 70 a;a= 1. 83 = 1. 8 m/s^2

Had the answer come out negative it would have indicated that we had chosen the direction of the acceleration
incorrectly. It is also helpful to note that the tensions,Tand−T, add to zero. The tensionsTand−Tare a Newton’s
Third Law (N3L) pair of forces. Such a pair cannot act on the same object. The object in this case is the system.
From the point of view of the system, a Newton’s Third Law pair of forces is considered internal to the system.
Internal forces must always add to zero when a simultaneous set of Newton’s Second Law equations are solved. The
only forces remaining (on the left side of the equation) are the “external” forces acting on the system.

Findingthetensionintherope

Since the acceleration is known, we can substitute its value in either the Newton’s Second Law equation for the mass
mor the massM. Using the equation for the massmis a bit simpler so:

T= ( 30 )( 9. 8 )−( 30 )( 1. 83 ) = 239. 1 = 240 N

As a check for reasonableness we can determine if the tension is greater thanMgsinθand less thanmg. Try it and
see!

Check Your Understanding

1. True or False: In order for a hanging massM 1 to move a massM 2 up along an inclined plane,M 1 must be greater
   thanM 2.

Answer:False. As we saw in the last problem,mwas smaller thanMandMstill moved up the plane. There is,
however, a “special” value of the massmwhich can keep the system stationary or have the system move up or down
the plane with constant velocity. We consider this situation in question 2.

2. For a given massM, what value of the massmwould keep the system at rest or moving with a constant velocity?

Answer:We go back to the equations formandMin the last problem and assume the acceleration is zero.
Simplifying, we have:

mg−Mgsinθ= 0

Thusm=Msinθis the condition for which the system remains at rest or moves with constant velocity. A massm
of 16.9 kg would satisfy this condition in the last problem. (Can you see why?)

A Pulley System with Friction along the Inclined Plane

Illustrative Example

Refer back toFigure5.16. Assume that friction exists between the massM= 40 kgand the plane. A diagram of this
situation is identical toFigure5.16 except that friction is included opposing the motion ofM. (SeeFigure5.16).