A block of weight W = 500 N is supported by a cable CD, which in turn is suspended by cables AC and BC. Determine the required tension forces TCA and TCB so that the resultant force at point C equals zero.

Question

A block of weight [latex]W[/latex] = 500 [latex]N[/latex] is supported by a cable [latex]CD[/latex], which in turn is suspended by cables [latex]AC[/latex] and [latex]BC[/latex]. Determine the required tension forces [latex]T_{CA} [/latex] and [latex]T_{CB} [/latex] so that the resultant force at point [latex]C[/latex] equals zero.

Accepted Answer

Step 1: Because all cables are concurrent at point [latex]C[/latex], isolate point [latex]C[/latex] and show all forces.

Directions for forces [latex]T_{CD} [/latex], [latex]T_{CA} [/latex], and [latex]T_{CB} [/latex] are known, but their magnitudes are unknown. [latex]T_{CD} [/latex] is equal to the weight [latex]W[/latex], which acts vertically downward.

Step 2: Resolve each force into its x and y components.

[latex]T_{CA_{x} } =\frac{3}{5} T_{CA }[/latex]

[latex]T_{CA_{y} } =\frac{4}{5} T_{CA }[/latex]

[latex]T_{CB_{x} } = T_{CB } \cos 30^{\circ } =0.866 \space T_{CB }[/latex]

[latex]R_{CB_{y} } = T_{CB } \sin 30^{\circ } =0.5 \space T_{CB }[/latex]

[latex] T_{CD}=W=500 \space N[/latex]

Step 3: Write the resultant component force equations.

[latex]R_{x} =\sum{F_{x}} =T_{CA_{x} }+T_{CB_{x} }=0[/latex]

We set [latex]R_{x}[/latex] = 0 to comply with the condition that no resultant force develops at point [latex]C[/latex]; that is, all forces are balanced at this point.

[latex] herefore R_{x}=-\frac{3}{5} T_{CA }+0.866T_{CB }=0[/latex]

[latex] \frac{3}{5} T_{CA }=0.866T_{CB }[/latex] (1)

then,

[latex]R_{y} =\sum{F_{y}} =+T_{CA_{y} }+T_{CB_{y} }-500 \space N=0[/latex]

[latex] herefore R_{y} =\frac{4}{5} T_{CA }+ .5 T_{CB }-500 \space N=0[/latex]

[latex] \frac{4}{5} T_{CA }+0.5 T_{CB }=500 \space N [/latex] (2)

From writing the resultant equations [latex]T_{CA }[/latex] and [latex]T_{CB }[/latex], we obtain two equations, (1) [latex] \frac{3}{5} T_{CA }=0.866T_{CB }[/latex] and (2) [latex] \frac{4}{5} T_{CA }+0.5 T_{CB }=500 \space N [/latex] , containing two unknowns. Solving the two equations simultaneously,

[latex] \frac{3}{5} T_{CA }=0.866T_{CB }[/latex] (1)

[latex] \frac{4}{5} T_{CA }+0.5 T_{CB }=500 \space N [/latex] (2)

From equation (1) [latex] \frac{3}{5} T_{CA }=0.866T_{CB }[/latex] ,

[latex] T_{CA }= \frac{5}{3} (0.866\space T_{CB }) =1.44 \space T_{CB }[/latex]

Substituting into equation (2) [latex] \frac{4}{5} T_{CA }+0.5 T_{CB }=500 \space N [/latex] ,

[latex] \frac{4}{5}( 1.44 \space T_{CB})+0.5 \space T_{CB }=500 \space N[/latex]

[latex] herefore 1.15 \space T_{CB}+0.5 \space T_{CB }=500 \space N[/latex]

Solving for [latex]T_{CB}[/latex],

[latex]1.65\space T_{CB }=500 \space N[/latex]

[latex] T_{CB }=\frac{500 \space N}{1.65} = 303 \space N[/latex]

Substituting the value of [latex] T_{CB }[/latex] back into equation (1) [latex] \frac{3}{5} T_{CA }=0.866T_{CB }[/latex] or (2) [latex] \frac{4}{5} T_{CA }+0.5 T_{CB }=500 \space N [/latex] ,

[latex]T_{CA }= 436.4 \space N [/latex]

Graphical Check:

Because [latex]T_{CD }[/latex] has a known magnitude and direction, it will be used as the starting (or base) force

Step 1: Draw the force polygon using the tip-to-tail method.

Step 2: Draw force [latex]T_{CD }[/latex] first, to scale.

Step 3: Draw the lines of action for [latex]T_{CA }[/latex] and [latex]T_{CB }[/latex]; the order does not matter.

Forces [latex]T_{CA }[/latex] and [latex]T_{CB }[/latex] have known directions but unknown magnitudes; therefore, only their lines of action are drawn initially. We know that because [latex]R[/latex] = 0, the tip of the last force must end at the tail (the origin) of the first force—in this case, [latex]T_{CD }[/latex].

Step 4: The intersection of the two lines of action determines the limits for[latex]T_{CB }[/latex] and [latex]T_{CA }[/latex].

Step 5: Scale off the magnitudes for [latex]T_{CB }[/latex] and [latex]T_{CA }[/latex].

Step 6:

[latex] T_{CA }= 436.4 \space N[/latex]

[latex]\space \space \space \space \space \space \space \space \space \space \space \space [/latex] CHECKS

[latex] T_{CB }= 303 \space N[/latex]

Question 2.11: A block of weight W = 500 N is supported by a cable CD, whic...

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