FORCE AND ACCELERATION: What is the mass of the climbing acrobat?

Q: Two acrobats, a pulley, and a rope are used in a circus act. Acrobat 1 rapidly climbs one of the suspended lengths of rope at a distance of 16ft in 2 seconds with a constant acceleration. On the opposite length of rope, acrobat 2 is suspended in an attached chair that remains motionless above the ground. Acrobat 2 and the chair have a combined weight of ( F_w2 ). The gravitational constant of acceleration ( g ) is 32ft/s².

( a ) What is the weight of acrobat 1 in relation to ( F_w2 )? 

( b ) Why is it necessary for acrobat 1 to climb the rope with constant acceleration? 

( c ) What would happen if the acrobat climbed the rope with a constant velocity instead?

( d ) If we wanted acrobat 2 to descend with a constant velocity, how would this be accomplished?

( e ) If acrobat 2 weighed less than acrobat 1, at what rate would acrobat 1 have to accelerate downward to keep acrobat 2 in a stationary and suspended location above the ground? 

A: Within the aforementioned system, the only thing in motion is acrobat 1: 

Prior to answering part ( a ) of the question, let’s consider an analogous circumstance where acrobat 1 stands atop a motionless treadmill. The sum of all forces acting upon his system equals zero, and he is stationary with respect to the members of the audience. If the treadmill’s belt began to move with a constant velocity ( v_t ), the acrobat would begin to move with a velocity equal to ( v_t ) with respect to the stationary observers in the audience. Under these conditions, the acrobat will have zero unbalanced forces exerted upon him. From the acrobat’s vantage point, there is nothing physically different about standing on a motionless treadmill belt vs standing on a belt that moves at a constant velocity. If he wished, the acrobat could just as easily balance a glass of water atop his head as he could have while standing still atop the motionless treadmill belt, because his momentum ( p = mv ) doesn’t change. 

If the acrobat instead wants to maintain a motionless stance with respect to the audience as the treadmill continues moving with a velocity ( v_t ), he would need to begin walking in the direction that opposes the motion of the treadmill belt with a constant velocity ( – v_a = v_t ). Once again, the acrobat would have no trouble balancing a glass of water atop his head under this circumstance, because his momentum ( p = mv ) does not change. 

If the treadmill began to accelerate, the acrobat’s experience would be significantly different than it was beforehand; his velocity must now change at a steady rate to maintain a stationary stance relative to the members of the audience. Since F = ma = ( m )( Δv/Δt ), the net force of the system will remain zero as long as the momentum of the acrobat changes in proportion to the changing velocity of the treadmill belt. As soon as the acrobat can no longer increase his velocity at the same rate as the treadmill belt, an unbalanced force will be exerted on his shoes from beneath. If the acceleration of the treadmill is great enough, the acrobat could easily tumble off of the treadmill ( to the glee and amusement of the audience ). 

Atop an accelerating treadmill of infinite length, the acrobat could change his position within the accelerating system by moving with a constant velocity, but he would still accelerate with the system in the same manner that a flight attendant walking down an accelerating plane’s aisle continues accelerating with the plane relative to onlookers at an airport; thus, acrobat 1 must accelerate while climbing the rope in order to create a counteracting force that, along with his weight, will balance the weight of acrobat 2. If the acrobat climbed the rope with a constant velocity, he would change his position on the rope without affecting the rope’s change in momentum [ Δp = ( m )( Δv ) ].

Acrobat 1 has a mass ( m₁ ), and acrobat 2 has a mass ( m₂ ). Additionally, acrobat 1 has a weight ( F_w1 = m₁g ), and the weight of acrobat 2 is ( F_w2 = m₂g ). Recall that acrobat 1 climbed 16ft upward on the rope in 2 seconds. We determine the rate of acceleration of acrobat 1 as follows:

d =  v_ot + ½ at² 

Keep in mind that acrobat 1 accelerates upward in the positive direction, but the force he exerts on the rope opposes ( F_w2 )! Additionally, let’s assume that acrobat 1 has an initial velocity ( v_o ) = 0:

d = ½ at²

Substitution of known values in the equation above will enable us to determine the value of ( a ):

16ft = ½ ( a )( 2s )²

32ft = ( a )( 4s² )  

a = 32ft/4s² = 8ft/s²

The net forces within this system are as follows:

F_w2 – m₁a – m₁g = ma_t

ma_t = 0

In order to determine the value of F_w1, we must know the value of m₁:

F_w2 = m₁a + m₁g

F_w2 = ( m₁ )( a + g )

F_w2 = ( m₁ )( 8ft/s² + 32ft/s² )

F_w2 = ( m₁ )( 40ft/s² )

We don’t know the value of m₂, but we do know that F_w2 = m₂g:

m₂g = ( m₁ )( 40ft/s² )

( m₂ )[ ( 32ft/s² )/( 40ft/s² ) ] = m₁

( ⁴/₅ )( m₂ ) = m₁

⅘ F₂ = F₁

Since the mass of acrobat 1 is ⅘ the mass acrobat 2, and F_w = mg, acrobat 1 must weigh ⅘ as much as acrobat 2 sitting in the suspended chair.

If we wanted acrobat 2 to descend with a constant velocity, how would this be accomplished?

Whenever the net force acting upon a system is zero, the system is either sitting still or moving with a constant velocity with respect to an observer. In order for acrobat 2 to descend with a constant velocity, the system must be allowed to accelerate until the desired velocity is achieved. At that point, acrobat 1 would need to begin accelerating up the rope at 8ft/s².

Finally, if acrobat 1 weighed more than acrobat 2, at what rate would acrobat 1 have to accelerate downward to keep acrobat 2 in a stationary and suspended location above the ground?

If F_w1 > F_w2, the weight of acrobat 1 would cause acrobat 2 to accelerate upward. If acrobat 1 completely releases the rope, acrobat 2 would fall towards the ground. For this reason, acrobat 1 must find a way to push upward upon the rope in order to bring the rise of acrobat 2 to a halt. This is accomplished by climbing downward with constant acceleration ( a ). By accelerating down the rope, acrobat 1 effectively counteracts his body weight’s ability to pull acrobat 2 upward:

– F_w1 + m₁a + m₂g = 0

– m₁g + m₁a + m₂g = 0

( m₁ )( a – g ) = – m₂g

( m₁ ) = ( – m₂ )[ g/( a – g ) ]

Note: In the former example, ( m₁ ) = ( m₂ )[ g/( a + g ) ].

Whatever the value of a happens to be, it cannot be greater than or equal to ( + g ) in the first example and ( – g ) in the latter case. A value of a > g would give us a ( – m₁ ) value, which would be incorrect. A value of a = g describes a circumstance where ( m₁ ) is floating weightlessly in mid-air. If ( a ) = 0, m₁ and m₂ would have to be equal for the net force of the system to equal zero. Any value of ( – a ) would represent a system where m₂ > m₁, and F_w2 > F_w1, with acceleration being in the reverse direction, as was the case in the previous example where ( F_w2 – m₁a – m₁g = 0 ). If ( a ) = 8ft/s², we’d obtain the following result:

( m₁ ) = ( – m₂ )[ 32ft/s²/( 8ft/s² – 32ft/s² ) ]

( m₁ ) = ( – m₂ )[ 32ft/s²/( -24ft/s² ) ]

( -24ft/s²/32ft/s² )( m₁ ) = ( – m₂ )

( ¾ )( m₁ ) = m₂

( ¾ )( F_w1 ) = ( F_w2 )

( F_w1 ) > ( F_w2 )

In the former circumstance, the larger mass ( m₂ ) is 20% heavier than ( m₁ ):

 m₂ > m₁

⅘ F₂ = F₁ 

In the latter circumstance immediately above, the larger mass ( m₁ ) is 25% heavier than ( m₂ ):

m₁ > m₂

¾ F_w1 = F_w2 

The same acceleration ( 8ft/s² ) was used to establish balance within both systems. The 5% increased ability of an acrobat to establish static equilibrium in the latter circumstance is due to the acrobat pushing upwards on the rope vs pulling downward. In the circumstance where he must pull the rope downward, he pulls himself upwards against the force of gravity into a higher potential energy state within the gravitational field. In the latter circumstance, the acrobat works with the force of gravity when he pushes himself downward and establishes the initial change in momentum needed to establish systemic equilibrium.