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Wize University Statics Textbook (Master) > Equilibrium of a Particle

(2D) Pulleys, Ropes, Cables, Springs

Free-Body / Force-Body Diagrams (FBD)Previous Section
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Ropes and Cables

Ropes and cables typically connect different parts of a system, potentially around a pulley. There are a few important concepts about them:
  • We assume ropes do not stretch
  • Tension is the same throughout the same rope
  • Ropes have negligible mass
  • Ropes are ALWAYS IN TENSION
Draw FBD for Ropes - Cables.

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Springs

Springs are also commonly seen in this course. You need to know the following about springs:
  • Springs are assumed to obey Hooke’s law
F=ksF=ks
s=l−lo,s=l-l_{o,}
  • The force in the spring is proportional to the stretch in the spring from its equilibrium position
  • Springs can be in tension or compression.


Spring in Action.



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Pulleys

-A pulley consists of a wheel with a groove and a cable/rope going around it.
-The pulleys we will analyze are FRICTIONLESS PULLEYS - can rotate freely, with negligible mass, without friction.


At Equilibrium,


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Examples




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Example

If the tension in BC is 500N, determine the tension in BD and AB, and determine the stretch in the spring.





Solution:

1) Draw Free Body Diagram




2) Apply Equation of Equilibrium

ΣFy=0\Sigma F_y=0
500 sin⁡45−FBDsin⁡36.9=0500\ \sin45-F_{BD}\sin36.9=0
FBD=589NF_{BD}=589N

ΣFx=0\Sigma F_x=0
500cos⁡45+589cos⁡36.9=Fs500\cos45+589\cos36.9=F_s
FAB=Fs=825NF_{AB}=F_s=825N

3) Solve for
Fs=k sF_s=k\ s
s=Fsk=825N800 N/ms=\frac{F_s}{k}=\frac{825N}{800\ N/m}
s=1.03ms=1.03m

Practice

Determine the mass of A needed to hold the system at equilibrium.





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Example

A guy with a mass of 90 kg supports himself in the bosun chair by means of the cable pulley system shown. Assuming the seat has a mass of 8kg, determine the tension force, T that the guy must exert with his hand on the cable at A in order the maintain equilibrium. Also, what force R, does the worker exert on the chair.
1) Label and determine where the tension is the same.

2) Begin drawing Free Body Diagrams (Assuming all cables are vertical)

We realize the tension T1 is what we solving for in the first step and this would have to depend on T2, therefore it is necessary for us to get T2 in terms of T1 as shown below.

Draw FBDs from the bottom up, and apply equilibrium equation ,

FBD (SEAT)


+↑ΣFy=0:T2−R−Wseat =0R=T2−Wseat R=T2−78.48N\begin{array}{l}{+\uparrow \Sigma F_{y}=0 : T_{2}-R-W_{\text {seat }}=0} \\ {R=T_{2}-W_{\text {seat }}} \\ {R=T_{2}-78.48 \mathrm{N}}\end{array}


FBD (GUY)

+↑ΣFy=0:T1+R−Wguy=0T1=Wguy−RT1=882.9N−R\begin{array}{l}{+\uparrow \Sigma F_{y}=0 : T_{1}+R-W_{g u y}=0} \\ {T_{1}=W_{g u y}-R} \\ {T_{1}=882.9 \mathrm{N}-R}\end{array}

FBD (B)

+↑ΣFy=0:−T2+3T1=03T1=T2T2=3T1\begin{array}{l}{+\uparrow \Sigma F_{y}=0 :-T_{2}+3 T_{1}=0} \\ {3 T_{1}=T_{2}} \\ {T_{2}=3 T_{1}}\end{array}


Substitute R=T2−78.48NR=T_{2}-78.48 N from FBD(SEAT) into R for equation T1=882.9N−RT_{1}=882.9 N-R from FBD(GUY).

Therefore,

T1=882.9N−T2+78.48NT_{1}=882.9 N-T_{2}+78.48 N

We can use Equation T2=3T1T_{2}=3 T_{1}from FBD(B) which relates T2 to T1 and plug it in to the above equation.

Therefore,

T1=882.9N−(3T1)+78.48NT_{1}=882.9 N-\left(3 T_{1}\right)+78.48 N

Solve for T1,

T1=T=240NT_{1}=T=240 N

Solving for R, from equation for FBD(GUY)

T1=882.9N−RR=882.9N−T1R=882.9N−240N=643N\begin{array}{l}{T_{1}=882.9 N-R} \\ {R=882.9 N-T_{1}} \\ {R=882.9 N-240 N=643 N}\end{array}
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Example

A guy with a mass of 90 kg supports himself in the bosun chair by means of the cable pulley system shown. Assuming the seat has a mass of 8kg, determine the tension force, T that the guy must exert with his hand on the cable at A in order the maintain equilibrium. Also, what force R, does the worker exert on the chair.
1) Label and determine where the tension is the same.

2) Begin drawing Free Body Diagrams (Assuming all cables are vertical)

We realize the tension T1 is what we solving for in the first step and this would have to depend on T2, therefore it is necessary for us to get T2 in terms of T1 as shown below.

Draw FBDs from the bottom up, and apply equilibrium equation ,

FBD (SEAT)


+↑ΣFy=0:T2−R−Wseat =0R=T2−Wseat R=T2−78.48N\begin{array}{l}{+\uparrow \Sigma F_{y}=0 : T_{2}-R-W_{\text {seat }}=0} \\ {R=T_{2}-W_{\text {seat }}} \\ {R=T_{2}-78.48 \mathrm{N}}\end{array}


FBD (GUY)

+↑ΣFy=0:T1+R−Wguy=0T1=Wguy−RT1=882.9N−R\begin{array}{l}{+\uparrow \Sigma F_{y}=0 : T_{1}+R-W_{g u y}=0} \\ {T_{1}=W_{g u y}-R} \\ {T_{1}=882.9 \mathrm{N}-R}\end{array}

FBD (B)

+↑ΣFy=0:−T2+3T1=03T1=T2T2=3T1\begin{array}{l}{+\uparrow \Sigma F_{y}=0 :-T_{2}+3 T_{1}=0} \\ {3 T_{1}=T_{2}} \\ {T_{2}=3 T_{1}}\end{array}


Substitute R=T2−78.48NR=T_{2}-78.48 N from FBD(SEAT) into R for equation T1=882.9N−RT_{1}=882.9 N-R from FBD(GUY).

Therefore,

T1=882.9N−T2+78.48NT_{1}=882.9 N-T_{2}+78.48 N

We can use Equation T2=3T1T_{2}=3 T_{1}from FBD(B) which relates T2 to T1 and plug it in to the above equation.

Therefore,

T1=882.9N−(3T1)+78.48NT_{1}=882.9 N-\left(3 T_{1}\right)+78.48 N

Solve for T1,

T1=T=240NT_{1}=T=240 N

Solving for R, from equation for FBD(GUY)

T1=882.9N−RR=882.9N−T1R=882.9N−240N=643N\begin{array}{l}{T_{1}=882.9 N-R} \\ {R=882.9 N-T_{1}} \\ {R=882.9 N-240 N=643 N}\end{array}

Practice

For the following pulley system if the maximum tension in each cable is 1500 N, determine the maximum weight of the platform under the following condition.

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If the tension in BC is 500N, determine the tension in BD and AB, and determine
the stretch in the spring.





Solution:


ΣFy=0\Sigma F_y=0
500 sin⁡45−FBDsin⁡36.9=0500\ \sin45-F_{BD}\sin36.9=0
FBD=589NF_{BD}=589N

Fs=kΔlF_s=k\Delta l ΣFx=0\Sigma{F}_x=0
Δl=Fsk=825N800 N/m\Delta l=\frac{F_s}{k}=\frac{825N}{800\ N/m} 500cos45+589cos36.9=Fs500cos45+589cos36.9=F_s
Δl=1.03m\Delta l=1.03m Fs=825NF_s=825N
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Determine the mass of A needed to hold the system at equilibrium.




Solution:
TBC=40g=392.4 NT_{BC}=40g=392.4\ N

ΣFy=0\Sigma{F}_y=0
392.4sin30−TAE=0392.4sin30-T_{AE}=0
TAE=196.2NT_{AE}=196.2N
TAE=MAgT_{AE}=M_Ag
196.2N=MA(9.81m/S2)196.2N=M_A(9.81m/S2)
MA=20kgM_A=20kg
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If l = 3ft, determine the tension each cable.

Solution:
2sin⁡θ=3sin⁡40 → θ=25.37o\frac{2}{\sin\theta}=\frac{3}{\sin40}\ \rightarrow\ \theta=25.37^o

ΣFx=0\Sigma {F}_x=0
TABcos40−TACcos25.37=0T_{AB}cos40-T_{AC}cos25.37=0
TAC=TABcos⁡40cos⁡25.37  (1)T_{AC}=T_{AB}\frac{\cos40}{\cos25.37}\ \ \left(1\right)

ΣFy=0\Sigma F_y=0
TAC sin⁡25.37+TABsin⁡40−200=0T_{AC}\ \sin25.37+T_{AB}\sin40-200=0
(cos⁡40cos⁡25.37sin⁡25.3+sin⁡40)TAB=200\left(\frac{\cos40}{\cos25.37}\sin25.3+\sin40\right)T_{AB}=200
TAB=198.8NT_{AB}=198.8N
Sub into (1)

TAC=198.8cos40cos25.37T_{AC}=198.8\frac{cos40}{cos25.37}
TAC=168.5NT_{AC}=168.5N