Multiple-body systems

We can use Newton’s second law to analyze systems involving multiple bodies in much the same way that we use it for single bodies. We will usually consider bodies that are connected together with massless strings that do not stretch. This allows us to assume that each body has the same acceleration. The equation we use for multiple-body systems is

F⃗net=mtota⃗sys\displaystyle \vec{F}_{\text{net}} = m_{\text{tot}} \vec{a}_{\text{sys}},

where F⃗net\displaystyle \vec{F}_{\text{net}} is the vector sum of all forces acting on the system (not internal tension forces), mtot\displaystyle m_{\text{tot}} is sum of the masses of all bodies, and a⃗sys\displaystyle \vec{a}_{\text{sys}} is the acceleration of the system.

This is very straightforward when all bodies are in a line. Once we find the acceleration, we can determine the force of tension acting on individual bodies using Newton’s second law again, but this time considering the body individually (draw a new free body diagram) and having acceleration as a known variable.

Pulley systems look two-dimensional, but we can turn them into one-dimensional problems by making one direction of the pulley negative and the other positive. This means, for example, that the force of gravity could be positive for one body and negative for another if they are on opposite sides of the pulley.

Example

In a pulley system where mass A (21 kg) is on a 18º frictionless ramp and mass B (12 kg) is suspended, find the acceleration of the system and the tension in the string.

As always, we start with a sketch:

18ºAB[bwd][fwd]
A pulley system of two bodies

There are three forces acting on the system:

Adding the first two gives us the parallel component of F⃗g,A\displaystyle \vec{F}_{\text{g,A}}:

F⃗g,A+F⃗N,A=F⃗g,ramp=∣F⃗g,A∣cos⁡(90º−θ)=∣F⃗g,A∣sin⁡θ\displaystyle \vec{F}_{\text{g,A}} + \vec{F}_{\text{N,A}} = \vec{F}_{\text{g,ramp}} = | \vec{F}_{\text{g,A}} \left \lvert \cos{\left ( 90 º - \theta \right )} = \mid \vec{F}_{\text{g,A}} \right \rvert \sin{\theta}.

Now we can use Newton’s second law to find acceleration.

F⃗net=∣F⃗g,A∣sin⁡θ+F⃗g,B=mtota⃗sys\displaystyle \vec{F}_{\text{net}} = \left \lvert \vec{F}_{\text{g,A}} \right \rvert \sin{\theta} + \vec{F}_{\text{g,B}} = m_{\text{tot}} \vec{a}_{\text{sys}}.

If we let [fwd] be positive, then F⃗g,ramp\displaystyle \vec{F}_{\text{g,ramp}} will be negative. Since F⃗g=mg\displaystyle \vec{F}_{\text{g}} = m g,

F⃗net=−mAgsin⁡θ+mBg=mtota⃗sys\displaystyle \vec{F}_{\text{net}} = - m_{\text{A}} g{\sin{\theta}} + m_{\text{B}} g{=} m_{\text{tot}} \vec{a}_{\text{sys}}.

Rearranging to solve for acceleration gives us

a⃗sys=−mAgsin⁡θ+mBgmtot\displaystyle \vec{a}_{\text{sys}} = \frac{- m_{\text{A}} g{\sin{\theta}} + m_{\text{B}} g}{m_{\text{tot}}},

and after substituting we have

a⃗sys=−(21 kg)(9.80 m/s2)sin⁡18º+(12 kg)(9.80 m/s2)21 kg+12 kg\displaystyle \vec{a}_{\text{sys}} = \frac{- \left ( 21 \, \text{kg} \right ) \left ( 9.80 \, \text{m/s}^{2} \right ) \sin{18} º + \left ( 12 \, \text{kg} \right ) \left ( 9.80 \, \text{m/s}^{2} \right )}{21 \, \text{kg} + 12 \, \text{kg}},

which evaluates to 1.6 m/s2 [fwd].

Now we can use the acceleration to find the tension in the string. We can do this using either A or B. It’s slightly easier with B. There are two forces acting on B: gravity and tension. We can write our net force equation again:

F⃗net=F⃗g,B+F⃗T=mBg+F⃗T=mBa⃗\displaystyle \vec{F}_{\text{net}} = \vec{F}_{\text{g,B}} + \vec{F}_{\text{T}} = m_{\text{B}} g + \vec{F}_{\text{T}} = m_{\text{B}} \vec{a}.

Solving for the force of tension, we have

F⃗T=mBa⃗−mBg=mB(a⃗−g)\displaystyle \vec{F}_{\text{T}} = m_{\text{B}} \vec{a} - m_{\text{B}} g{=} m_{\text{B}} \left ( \vec{a} - g \right ),

which gives us the solution

F⃗T=(12 kg)(1.64 m/s2−9.80 m/s2)=−97.92 N\displaystyle \vec{F}_{\text{T}} = \left ( 12 \, \text{kg} \right ) \left ( 1.64 \, \text{m/s}^{2} - 9.80 \, \text{m/s}^{2} \right ) = - 97.92 \, \text{N},

therefore the magnitude of the force of tension is 98 N.