Q: Three objects are connected by ropes that pass over massless and frictionless pulleys. As the objects move, the table exerts a force of friction on the middle object. The coefficient of kinetic friction is 0.100.
- What is the acceleration of the three objects within the system?
- What is the magnitude of the tension in each rope?
A: Newton’s Second Law of Motion states that the rate of change in momentum of an object ( acceleration ) is directly proportional to an applied net force ( Fnet = ma ). Newton’s Third Law of Motion states that all forces occur in pairs. If a force acting upon an object is opposed by an equal and oppositely oriented force, the object will ( a ) remain at rest, or ( b ) move with a constant velocity ( a = 0 ). The SI unit of force is the Newton ( N ), and the magnitude of a force is the product of mass in kilograms ( kg ) and acceleration, which has subunits of ( kg*m/s2 ).
In the diagram below, three objects with different masses are attached via two ropes. Because of this attachment, the objects will all accelerate at the same rate; however, since F = ma, and the masses of each object is different, the forces acting upon the objects will differ. Each rope represents a force pair, which is referred to as tension ( T ). Tension is equally distributed throughout each rope. Furthermore, we must account for the force of friction ( Ff ) acting upon the middle object. The force of friction always opposes the direction of motion:
Since the weight ( Fw ) of the suspended objects acts downward, the more massive object to the right will determine the direction in which the objects within the system accelerate:
Fw = mg, and the value of the gravitational constant of acceleration ( g ) is 9.8 m/s2:
Prior to determining how the tension in each rope will influence the motion of the system, it will be convenient to determine what forces act upon the object at the center of the system. The normal force ( N ) acting upward on the object is part of a force pair; the force equal and opposite to the normal force is the weight of the object. Because N = mg, the middle object accelerates horizontally, but its vertical acceleration is zero. Furthermore, the force of friction opposing its motion is ( Ff ) = uN:
The diagrammatic representation of the system will be complete when the tension forces in each rope are taken into consideration:
With all forces taken into account, we could use free-body diagrams to analyze the net forces ( Fnet ) acting upon each object. Since a sketch of the system has already been set up, free-body diagrams will be omitted for convenience. The net forces acting upon the first object are as follows:
Fnet = m1a
T1 – m1g = m1a
T2 – T1 – um2g = m2a
m3g – T2 = m3a
We now have three equations that, when added together, yield the following result:
m3g – um2g – m1g = m1a + m2a + m3a
( g )( m3 – um2 – m1 ) = ( a )( m1 + m2 + m3 )
a = ( g )[ ( m3 – um2 – m1 ) / ( m1 + m2 + m3 ) ]
An alternative ( and more difficult ) method of solving the problem is as follows:
T2 = m2a + um2g + T1
T2 = m3g – m3a
m2a + um2g + T1 = m3g – m3a
m2a + m3a + T1 = m3g – um2g
We must now solve T1 – m1g = m1a for T1 and substitute it into the equation above:
T1 = m1a + m1g
m2a + m3a + m1a + m1g = m3g – um2g
m2a + m3a + m1a = m3g – um2g – m1g
( a )( m1 + m2 + m3 ) = ( g )( m3 – um2 – m1 )
a = ( g )[ ( m3 – um2 – m1 ) / ( m1 + m2 + m3 ) ]
The [ ( m3 – um2 – m1 ) / ( m1 + m2 + m3 ) ] term from the equation above shows that ( a ) will be some fraction of ( g ). This makes sense, because the ( g ) represents the acceleration of an object in free fall near the earth’s surface. Substitution of ( a ) into T1 = m1a + m1g allows us to determine the value of T1. Likewise, substitution of ( a ) into T2 = m3g – m3a will allow us to determine the value of T2.
RENAISSANCE PHYSICS 