Question

Why doesn't this problem work using trigonometic functions?

For example if you try cos(30) = 600lbs/Fu you will not get the right answer.

What is the rule so you know you HAVE to use law of sines?

Please explain in words, DO NOT rework the problem. The problem is already worked out.

Resolve the horizontal 600-lb force in Fig. 2-12a into components acting along the u and axes and determine the magnitudes of these componcnts. 30 120 120 600 lb 600 1b 600 Ib Fig. 2-12 SOLUTION The parallelogram is constructed by extending a line from the head of the 600-lb force parallel to the vaxis until it intersects the u axis at point B. Fig. 2-12b. The arrow from A to B rcprcsents F. Similarly the line extended from the head of the 600-lb force drawn parallel to the u axis intersccts the v axis at point C, which gives F The vector addition using the triangle rule is shown in Fig. 2-12c. The two unknowns are thc magnitudes of Fu and F Applying the law of ines, 600 lb sin 120 sin 30° Fu 1039 lb 600 lb is sin 30°sin 30 F600 Ib Ans NOTE: The result for Fu shows that sometimes a component can have a grcatcr magnitude than thc resultant.

Answer 1