Two cables are tied together at C and loaded as shown. Assuming that the maximum permissible tension in CA and CB is 3 kN (the safe capacity of the cable), determine the maximum W that can be safely supported.

Question

Two cables are tied together at [latex]C[/latex] and loaded as shown. Assuming that the maximum permissible tension in [latex]CA[/latex] and [latex]CB[/latex] is [latex]3 kN[/latex] (the safe capacity of the cable), determine the maximum [latex]W[/latex] that can be safely supported.

Accepted Answer

Force	[latex]F_{x}[/latex]	[latex]F_{y}[/latex]
CA	[latex]-\frac{5}{13} [/latex]CA	[latex]+\frac{12}{13} [/latex]CA
CB	[latex]+\frac{4}{5} [/latex]CB	[latex]+\frac{3}{5} [/latex]CB
W	0	-W

[latex]\left[\sum{F_{x}}=0 ight] =-\frac{5}{13} CA + \frac{4}{5} CB=0[/latex] (1)

[latex] CA =\frac{13}{5} imes \frac{4}{5} CB= \frac{52}{25} CB[/latex]

[latex] herefore CA = 2.08 \space CB[/latex]

This relationship is crucial, because it tells us that for the given arrangement of cables, [latex]CA[/latex] will carry more than twice the load in [latex]CB[/latex] for equilibrium to exist. Because the maximum cable tension is restricted to [latex]3 kN[/latex], this value should be assigned to the larger of the two cable tensions.

[latex] herefore CA = 3 \space kN[/latex]

[latex]CB= \frac{CA}{2.08}=\frac{ 3 \space kN}{2.08}=1.44 \space kN[/latex]

Note: If the [latex]3 kN[/latex] value were assigned to [latex]CB[/latex], then [latex]CA[/latex] would be [latex]6.24 kN[/latex], which obviously exceeds the allowable cable capacity.

This problem is completed by writing the second equation of equilibrium.

[latex]\left[\sum{F_{y}}=0 ight] +\frac{12}{13} CA + \frac{3}{5} CB - W=0[/latex] (2)

Substituting the values for [latex]CA[/latex] and [latex]CB[/latex] into equation (2) [latex]\left[\sum{F_{y}}=0 ight] +\frac{12}{13} CA + \frac{3}{5} CB - W =0[/latex] ,

[latex]+\frac{12}{13} (3 \space kN) + \frac{3}{5} (1.44 \space kN )=W[/latex]

[latex] W=2.77 \space kN + 0.86 \space kN =3.63 \space kN[/latex]

Graphical Solution (Scale: [latex]1 mm = 50 N[/latex]):

Using the tip-to-tail method, a force triangle is constructed such that [latex]W[/latex], [latex]CA[/latex], and [latex]CB[/latex] form a closed triangle. The tip of the last force must end on the tail of the first force for

[latex]R_{x} =\sum{F_{x}} =0[/latex] and [latex]R_{y} =\sum{F_{y}} =0[/latex]

Weight [latex]W[/latex] is known to be a vertical force that closes at the origin [latex]O[/latex]. The only thing known about [latex]CB[/latex] and [latex]CA[/latex] is their lines of action. Visually, however, it is apparent that [latex]CA[/latex] must be the [latex]3 kN[/latex] tension force so that [latex]CB[/latex] does not exceed the allowable tension. If [latex]CB[/latex] were drawn as [latex]3 kN[/latex] , [latex]CA[/latex] would end up being much larger.

Question 2.24: Two cables are tied together at C and loaded as shown. Assum...

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