1. Define the Unit System and Coordinate System

Example Problem 3-6

The cable system, shown in fig (a), is being used to lift body E. The cable system is in equilibrium at the cable positions shown when a 500-N force is applied at a joint 1. Determine the tensions in cables A, B, C and D, and the mass of a body E that is being lifted.

Steps:

1.  Define the unit system and coordinate system.

2.  Identify the appropriate system(s).

3.  Apply the conservation principle.

1. Since the problem is given in metric units, we choose the unit system shown above. A Cartesian coordinate system is chosen because it is the most convenient for this problem.

2. The choice of an appropriate system (or systems) is extremely important. In this case, we look at the problem sketch and see that we have the following unknown quantities, tensions in cables A, B, C and D, and the mass of the body E that is being lifted. We define these tensions to be TA, TB, TC and TD. Note that there are 3 joints (labeled 1, 2 and 3). On the sketch to the right, four "systems" are shown by the dotted lines. In each case, we have chosen the system to be one or more joints. For the large system that includes joints 1 and 2, we note that the tension in cables C, B and D are "cut" by the system boundary. The free-body diagram for this system will include tensions TB, TC and TD as shown below.

We note that the 3 tensions TB, TC and TD are unknown. Since the problem is planar in the x-y plane, we can write conservation of linear momentum in the x and y directions (2 equations). However, conservation in the z direction will produce a trivial (0=0) equation. Consequently, this system is not a good choice. Suppose we choose the system to include only one joint (joint 1, 2 or 3). The resulting free-body diagrams are shown below. Note that cable A connects joints 1 and 2. The tension in cable A (TA) therefore acts on joints 1 and 2 and is drawn out of each joint (or pulling on the joint in tension), and is so drawn consistently on all free-body diagrams that involve TA. Note how TA act "equal and opposite directions" on joints 1 and 2. A similar situation occurs for TD, which acts on joints 2 and 3. It is extremely important that the vector directions for a given tension be drawn correctly on each free-body.

In the free-body diagram for joint 1, we see that there are two unknown tensions: TA and TC. The free-body diagram for joint 2 contains three unknown tensions: TA , TB , and TD. We will see that if we start with the free-body diagram for joint 2, we will be able to determine only 2 non-trivial conservation equations. Hence, it would be wise to start with the free-body diagram (system) for joint 1 and solve for tensions TA and TC. Once TA is obtained, we can use the free-body for joint 2 and solve for TB and TD. Then the free-body diagram for joint 3 can be used to solve for the weight WE.

We can define the tensions in terms of Cartesian vector form. As discussed above, we will consider each free-body diagram (system) separately (but in the correct order) and apply conservation of linear momentum (rate equation) to each of these.

3. Apply Conservation of Linear Momentum to each free-body (system).

Considering the free-body in fig. (b):

TA = TA (1 i + 0 j + 0 k) N

TC = TC (- cos 10o i + sin 10o j + 0 k) N

F1 = F1 (-j); where F1=500 N

Now we apply the conservation of linear momentum (rate equation).

Conservation of Linear Momentum (LM) ; Rate Equation / / = / / + /
Accum rate of the system LM / = / Input/Output rate of LM due to mass in/out / + / Input/Output rate of LM due to external forces
Units / Kg m/s2 / Kg m/s2 / Kg m/s2
TA / TC / F1
x-direction, / 0 / = / 0 / + / TA / + / -TC cos 10o / + / 0
y-direction, / 0 / = / 0 / + / 0 / + / TC sin 10o / + / -500
z-direction, / 0 / = / 0 / + / 0 / + / 0 / + / 0

From the above conservation table, we can solve the first two equations to obtain the two unknowns, namely TA and TC. Note that the z-direction produces a trivial equation (0=0).

We obtain two equations from COLM:

TA + (-TC cos 10o) = 0 ------(1)

TC sin 10o +(-500) = 0 ------(2)

From equation (2 ), we obtain TC = (500)/ (sin 10o)

Therefore, TC = 2,880 N

Substituting the value of TC = 2,880 N into eqn. (1), one obtains:

TA = TC cos 10o

\ TA = 2,880 cos 10o N

\ TA = 2,840 N

Considering the free-body in fig. (c):

TA = TA ( -i + 0 j + 0 k) N

TB= TB ( sin 20o i+cos 20o j+0 k)N

TD =TD ( 0 i - 1 j+0 k) N

Now we apply the conservation of linear momentum (rate equation).

Conservation of Linear Momentum (LM) ; Rate Equation / / = / / + /
Accum rate of the system LM / = / Input/Output rate of LM due to mass in/out / + / Input/Output rate of LM due to external forces
Units / Kg m/s2 / Kg m/s2 / Kg m/s2
TA / TB / TD
x-direction, / 0 / = / 0 / + / -TA / + / TB sin 20o / + / 0
y-direction, / 0 / = / 0 / + / 0 / + / TB cos 20o / + / -TD
z-direction, / 0 / = / 0 / + / 0 / + / 0 / + / 0

From the above conservation table, we can solve the first two equations to obtain the two unknowns TB and TD. Note that once again we obtain no useful information from the z-direction conservation equation (0=0).

We get obtain two equations from COLM:

-TA + (TB sin 20o) = 0 ------(3)

TB cos 20o +(- TD) = 0 ------(4)

Substituting the value of TA =2,840 N into equation (3 ), we obtain,

TB = (TA)/ (sin 20o)

\ TB = (2,840)/(sin 20o) N

Or, TB = 8,290 N.

Substituting the value of TB = 8290 N. into eqn. (4), we obtain

TD =(TB cos 20o)

\ TD =(2,880 cos 20o)

\ TD = 7,789 N

Considering the free-body in fig. (d):

TD = TD (0 i + 1 j + 0 k) N

WE = WE (0 i - 1 j + 0 k) N

Now apply the conservation of linear momentum (rate equation).

Conservation of Linear Momentum (LM) ; Rate Equation / / = / / + /
Accum rate of the system LM / = / Input/Output rate of LM due to mass in/out / + / Input/Output rate of LM due to external forces
Units / Kg m/s2 / Kg m/s2 / Kg m/s2
TD / WE
x-direction, / 0 / = / 0 / + / 0 / + / 0
y-direction, / 0 / = / 0 / + / TD / + / -WE
z-direction, / 0 / = / 0 / + / 0 / + / 0

From the above conservation table, we can solve the single equation to obtain the unknown WE.

We obtain one non-trivial equation from COLM:

TD + (- WE) = 0 ------(5)

Therefore, WE = TD. By substituting the value of TD =7,789 N into eqn. (5), we can determine the value of WE.

\ WE = TD = 7,789 N. The mass of the object is ME= WE/g= 794 Kg.

Note: We could have alternately solved the above set of five non-trivial equations (equations 1-5) simultaneously using EES, Maple, or a calculator. Remember that in EES, you can enter the equations as they are written above (no need to rearrange variables, put in matrix notation etc.)

TA = 2,840 N, TB = 8,290 N, TC = 2,880 N, TD = 7,789 N, WE = 7,789 N

ME= WE/g= 794 Kg

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