document.write( "Question 1164371: The ends X and Y of an inextensible string 27m long are fixed at two points on the same horizontal line which are 20m apart. A particle of mass 7.5kg is suspended from a point P on the string 12m from X.
\n" ); document.write( "a) illustrate this information in a diagram
\n" ); document.write( "b) calculate, correct to two decimal places angle YXP and angle XYP
\n" ); document.write( "c) find, correct to the nearest hundredth, the magnitudes of the tensions in the string. [Take g=10 m/s2] \n" ); document.write( "
Algebra.Com's Answer #854203 by KMST(5345)\"\" \"About 
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The ends X and Y of an inextensible string 27m long are fixed at two points on the same horizontal line which are \"20m\" apart.
\n" ); document.write( "A particle of mass 7.5kg is suspended from a point P on the string 12m from X.
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\n" ); document.write( "As it is an inextensible string, it will not stretch and its total length will always be \"27m\" . If the point P is at \"12m\" from point X, point P will be at \"27m-12m=15m\" from point Y.
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\n" ); document.write( "a) illustrate this information in a diagram
\n" ); document.write( "The diagram will show the triangle, XYZ with the length shown as numbers without the units,
\n" ); document.write( "as it is understood that they the lengths are in meters.
\n" ); document.write( "For easy reference, the height \"green%28h%29\" with respect to base XY will be shown, as well as the length of \"green%28d%29\" of the projection of segment XP onto XY.
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\n" ); document.write( "b) calculate, correct to two decimal places angle YXP and angle XYP
\n" ); document.write( "The angles in question are those at X and Y. For short, those angles and their measures will be referred to as \"X\" and \"Y\" .
\n" ); document.write( "The measuring units are not specified, but will be calculated in degrees.
\n" ); document.write( "A first step would be calculating trigonometric functions of the angles, such as \"cos%28X%29\" and \"cos%28Y%29\" , from the known lengths.\r
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\n" ); document.write( "To calculate \"X\" and \"Y\" a student who has been taught how to solve triangles that are not necessarily right triangles could use the law of cosines to calculate \"cos%28X%29\" and \"cos%28Y%29\",
\n" ); document.write( "unless a different calculation is expected.
\n" ); document.write( "Law of cosines says that in a triangle ABC, with vertices A, B, and C opposite sides a, b, and c, respectively, the side and angle measures are related by
\n" ); document.write( "\"a%5E2=b%5E2%2Bc%5E2-2bc%2Acos%28A%29\" .
\n" ); document.write( "Applied to angle \"X\" , you get
\n" ); document.write( "\"15%5E2=20%5E2%2B12%5E2-2%2A20%2A12%2Acos%28X%29\"-->\"225=400%2B144-480%2Acos%28X%29\"-->\"225-400-144%2B480%2Acos%28X%29=0\"->\"-319%2B480%2Acos%28X%29=0\"->\"cos%28X%29=319%2F480=0.664583\"--> \"X=highlight%2848.35%5Eo%29\"
\n" ); document.write( "Applied to angle \"Y\" , you get
\n" ); document.write( "\"12%5E2=20%5E2%2B15%5E2-2%2A20%2A15%2Acos%28Y%29\"-->\"144=400%2B225-600%2Acos%28Y%29\"-->\"144-400-225%2B600%2Acos%28Y%29=0\"-->\"-481%2B600%2Acos%28Y%29=0\"->\"cos%28Y%29=481%2F600=0.801667\"--> \"Y=highlight%2836.71%5Eo%29\"
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\n" ); document.write( "For students who know Heron's (or Hero's) formula, they can calculate the area \"A\" of triangle XYZ, and from that and the length of side XY, calculate \"green%28h%29\" , and from there calculate \"sin%28X%29\" and \"sin%28Y%29\" to find \"X\" and \"Y\" .
\n" ); document.write( "Heron's (or Hero's) formula says that \"A=sqrt%28s%28s-a%29%28s-b%29%28s-c%29%29\" where a, b, and c are the lengths of the triangle sides, and \"s=%28a%2Bb%2Bc%29%2F2\" is the semiperimeter.
\n" ); document.write( "In this case \"s=%2820%2B15%2B12%29%2F2=47%2F2=23.5\"
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\n" ); document.write( "As the area of a triangle is \"base%2Aheight%2F2\" ,
\n" ); document.write( "\"89.6657=20%2Agreen%28h%29%2F2\"--> \"green%28h%29=V%2A2%2F20=8.96657%0D%0A%7B%7B%7Bsin%28X%29=green%28h%29%2F12=0.747214\"-->\"X=highlight%2848.35%5Eo%29\"
\n" ); document.write( "\"sin%28Y%29=green%28h%29%2F15=0.597771\"-->\"Y=highlight%2836.71%5Eo%29\"
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\n" ); document.write( "Another option is to consider the two right triangles formed by splitting triangle XYZ along the altitude labeled as \"green%28h%29\" , using the Pythagorean theorem to calculate \"green%28d%29\" , and calculating \"cos%28X%29\" and \"cos%28Y%29\" as trigonometric ratios.
\n" ); document.write( "The Pythagorean theorem only applies to right triangles. If A is the right angle opposite hypotenuse of length a, and the length of the legs of the right triangle are b and c,
\n" ); document.write( "then \"a%5E2=b%5E2%2Bc%5E2\" .
\n" ); document.write( "Applying it to the right triangle with side lengths (in meters) \"12\", \"green%28d%29\" and \"green%28h%29\", we get
\n" ); document.write( "\"12%5E2=green%28d%29%5E2%2Bgreen%28h%29%5E2\"-->\"highlight%28144-green%28d%29%5E2%29=green%28h%29%5E2\"
\n" ); document.write( "Applying it to the other right triangle, the one with side lengths (in meters) \"15\", \"20-green%28d%29\" and \"green%28h%29\", we get
\n" ); document.write( "\"15%5E2=%2820-green%28d%29%29%5E2%2Bgreen%28h%29%5E2\"-->\"225=400-2%2A20%2Agreen%28d%29%5E2%2Bgreen%28h%29%5E2\"-->\"225=400-40green%28d%29%5E2%2Bgreen%28h%29%5E2\"-->\"highlight%28225-400%2B40green%28d%29-green%28d%29%5E2%29=green%28h%29%5E2\"
\n" ); document.write( "As the two highlighted expressions are equal to \"green%28h%29%5E2\" , we can write
\n" ); document.write( "\"144-green%28d%29%5E2=225-400%2B40green%28d%29-green%28d%29%5E2%29\"-->\"144=225-400%2B40green%28d%29\"-->\"144-225%2B400=40green%28d%29\"-->\"319=40green%28d%29\"-->\"green%28d%29=319%2F40=7.975\"
\n" ); document.write( "\"cos%28X%29=green%28d%29%2F12=7.975%2F12=0.664583\"
\n" ); document.write( "\"cos%28Y%29=%2820-green%28d%29%29%2F15=%2820-7.975%29%2F15=12.025%2F15=0.801667\"\r
\n" ); document.write( "\n" ); document.write( "c) find, correct to the nearest hundredth, the magnitudes of the tensions in the string. [Take g=10 m/s2]
\n" ); document.write( "The weight of an object of mass \"m\" in a location with a gravitational acceleration \"g\" is a force \"W=m%2Ag\" .
\n" ); document.write( "With the mass in kg and the acceleration in \"%22m+%2F%22\"\"s%5E2\" , the resulting product will be the force in newtons, abbreviated \"N\" .
\n" ); document.write( "If we use \"g=10\"\"%22m+%2F%22\"\"s%5E2\" , the weight calculates as \"7.5%2A10N=75N\"
\n" ); document.write( "The tensions on both sides of the string are the forces pulling to balance the weight of
\n" ); document.write( "the 7.5kg particle. The three forces can be considered vectors that add up to zero force, meaning no net force on the particle, and no motion of the particle.
\n" ); document.write( "I’ll update the diagram from above with the proper angles, adding the three forces. I will call the tension to the left \"red%28L%29\" and the tension to the right \"red%28R%29\" .
\n" ); document.write( "I will use the same names to represent the numeric value of their magnitudes in newtons, without having to write the units at every step.\r
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\n" ); document.write( "A trick that works for many vector problems in Physics class is decomposing some them into a horizontal component and a vertical component. It works in this case.
\n" ); document.write( "To visualize horizontal and vertical component I drew right triangles that form a rectangular \"cages\" around \"red%28L%29\" and \"red%28R%29\" >
\n" ); document.write( "The magnitude of the horizontal component of a vector is the magnitude of the vector times the cosine of the smaller angle it forms with the horizontal.
\n" ); document.write( "The magnitude of the vertical component of a vector is the magnitude of the vector times the sine of the smaller angle it forms with the vertical.
\n" ); document.write( "I marked those angles with little green arcs showing that their measures are \"X\" and \"Y\" .
\n" ); document.write( "For the vectors \"red%28L%29\" , \"red%28R%29\" , and \"blue%28W%29\" to add to zero,
\n" ); document.write( "the vertical components of \"red%28L%29\" , \"red%28R%29\" must add to \"blue%28W%29\" ,
\n" ); document.write( "and the horizontal components of \"red%28L%29\" and \"red%28R%29\" must add to zero.
\n" ); document.write( "Those horizontal components point left and right, so their magnitudes must be equal. That means \"red%28L%29%2Acos%28X%29=red%28R%29%2Acos%28Y%29\"-->\"red%28L%29=red%28R%29%2Acos%28Y%29%2Fcos%28X%29\"-->\"red%28L%29=red%28R%29%2A0.801667%2F0.664583=1.20627red%28R%29\"
\n" ); document.write( "The magnitude of the vertical components adds to \"blue%28W%29=75\" , so
\n" ); document.write( "\"red%28L%29%2Asin%28X%29%2Bred%28R%29%2Asin%28Y%29=75\"
\n" ); document.write( "\"red%28L%29%2A0.74721%2Bred%28R%29%2A0.59777=75\"
\n" ); document.write( "\"1.20627red%28R%29%2A0.74721%2Bred%28R%29%2A0.59777=75\"
\n" ); document.write( "\"%281.20627%2A0.74721%2B0.5977%29%2Ared%28R%29=75\"-->\"1.49911%2Ared%28R%29=75\"-->\"red%28R%29=75%2F1.49911=highlight%2850.0298%29\"
\n" ); document.write( "\"red%28L%29=1.20627red%28R%29=1.20627%2A50.0298=highlight%2860.3494%29\"
\n" ); document.write( "I carried more than enough decimal places through the calculations,
\n" ); document.write( "but as the values for the mass an for g are given with just 2 significant figures,
\n" ); document.write( "I would report as results that the magnitudes of the tensions are
\n" ); document.write( "\"red%28L%29=highlight%2860N%29\" and \"red%28R%29=highlight%2850N%29\" , with just 2 significant figures. \n" ); document.write( "
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