PHY160-Solution-2H1-S24
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Feb 20, 2024
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PHY160-Solution-H2-Ch1-F12 SOLUTION There are six possible choices for the two vectors, leading to the following resultant vectors: F, + F, =50.0 newtons +10.0 newtons = +60.0 newtons = 60.0 newtons, due east F, + F; =50.0 newtons — 40.0 newtons = +10.0 newtons = 10.0 newtons, due east F, + F; = 50.0 newtons —30.0 newtons = +20.0 newtons = 20.0 newtons, due east Fz + F3 =10.0 newtons — 40.0 newtons = —30.0 newtons = 30.0 newtons, due west Fz - F4 =10.0 newtons — 30.0 newtons = —20.0 newtons = 20.0 newtons, due west F3 +F, =—40.0 newtons —30.0 newtons = —70.0 newtons = 70.0 newtons, due west The resultant vector with the smallest magnitude is IF, +F; =10.0 newtons, due east] : The resultant vector with the largest magnitude is |F3 +F, = 70.0 newtons, due west]. 24. 34. [SSM] REASONING AND SOLUTION In order to determine which vector has the largest x and y components, we calculate the magnitude of the x and y components explicitly and compare them. In the calculations, the symbol u denotes the units of the vectors. A, = (100.0 u) cos 90.0° = 0.00 u A), = (100.0 u) sin 90.0° = 1.00 x 10° u B, =(200.0 u) cos 60.0° = 1.00 x 10° u B‘, = (200.0 u) sin 60.0° =173 u C,=(150.0 u) cos 0.00° = 150.0 u C‘. =(150.0 u) sin 0.00° = 0.00 u a. [ C has the largest x component. | b. I B has the largest y component. I 36. REASONING Using trigonometry, we can determine the angle & from the relation tan @ = A4 /A,: SOLUTION &3 a. O=tan"" Y |=tan™! (12_m)= 45° | 12m 1 2m
38. @ REASONING The x and y components of r are mutually perpendicular; therefore, the magnitude of r can be found using the Pythagorean theorem. The direction of r can be found using the definition of the tangent function. SOLUTION According to the Pythagorean theorem, we have r=J-r2 '*'}’2 =\/(-125 m)* +(—184 m)* =I&l N/ The angle @ is o (184 m = e (125 sz 44 SOLUTION The x component of the resultant force F is F_=(2240 N)cos 34.0°+(3160 N)cos 90.0° = (2240 N)cos 34.0° F, Fy. Ax The y component of the resultant force F is F, =—(2240 N)sin 34.0°+(-3160 N) F F Ay By Using the Pythagorean theorem, we find that the magnitude of the resultant force is F=F +F =[(2240 N)cos 34.0°T +[—(2240 N)sin 34.0°—3160 N| =[ 4790 N | Using trigonometry, we find that the direction of the resultant force is o tant | (2240 N)sin 34.0°+ 3160 N (2240 N)cos 34.0° ]:l 67.2° south of east | 46: SOLUTION The first four rows of the table below give the components of the vectors __ A, B, C, and D. Note that east and north have been taken as the positive directions; hence vectors pointing due west and due south will appear with a negative sign. EastliWest NorthlSouth Vector Component Component A + 2.00 km 0 B 0 + 3.75 km C —2.50 km 0 D 0 —3.00 km R=A+B+C+D —0.50 km + 0.75 km The fifth row in the table gives the components of R. The magnitude of R is given by the Pythagorean theorem as R =./(—0.50 km)> +(+ 0.75 km)> =| 0.90 km | The angle @ that R makes with the direction due west is R ~1(0.75 km Ry & =tan \O.SOkm =|56° northofwest] Rp
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