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21_PartUniversity Physics Solution

# 21_PartUniversity Physics Solution - Units Physical...

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Units, Physical Quantities and Vectors 1-21 ( ) (0 148.5 m 242.5 m) 94.0 m y y y y D A B C = − + + = − + = − 2 2 x y D D D = + 2 2 ( 108.5 m) ( 94.0 m) 144 m D = + − = 94.0 m tan 0.8664 108.5 m y x D D θ = = = 180 40.9 220.9 θ = ° + ° = ° ( D " is in the third quadrant since both x D and y D are negative.) Figure 1.73b The direction of D " can also be specified in terms of 180 40.9 ; φ θ = ° = ° D " is 41 ° south of west. E VALUATE : The vector addition diagram, approximately to scale, is Vector D " in this diagram agrees qualitatively with our calculation using components. Figure 1.73c 1.74. I DENTIFY : Solve for one of the vectors in the vector sum. Use components. S ET U P : Use coordinates for which x + is east and y + is north. The vector displacements are: 2.00 km, 0 of east; 3.50 m, 45 south of east; = ° = ° A B \$ \$ and 5.80 m, 0 east = ° R \$ E XECUTE : ( ) ( )( ) 5.80 km 2.00 km 3.50 km cos45 1.33 km x x x x C R A B = = ° = ; y y y y C R A B = ( )( ) 0 km 0 km 3.50 km sin45 2.47 km = − − ° = ; ( ) ( ) 2 2 1.33 km 2.47 km 2.81 km C = + = ; ( ) ( ) 1 tan 2.47 km 1.33 km 61.7 north of east. θ = = ° The vector addition diagram in Figure 1.74 shows good qualitative agreement with these values. E VALUATE : The third leg lies in the first quadrant since its x and y components are both positive. Figure 1.74 1.75. I DENTIFY : The sum of the vector forces on the beam sum to zero, so their x components and their y components sum to zero. Solve for the components of F " . S ET U P : The forces on the beam are sketched in Figure 1.75a. Choose coordinates as shown in the sketch. The 100-N pull makes an angle of 30.0 40.0 70.0 + = ° ° ° with the horizontal. F " and the 100-N pull have been replaced by their x and y components. E XECUTE : (a) The sum of the x -components is equal to zero gives (100 N)cos70.0 0 x F + = ° and 34.2 N x F = − . The sum of the y -components is equal to zero gives (100 N)sin70.0 124 N 0 y F + = ° and 30.0 N y F = + . F " and its components are sketched in Figure 1.75b. 2 2 45.5 N x y F F F = + = . 30.0 N tan 34.2 N y x F F φ = = and 41.3 φ = ° . F " is directed at 41.3 ° above the x -axis in Figure 1.75a. (b) The vector addition diagram is given in Figure 1.75c. F " determined from the diagram agrees with F " calculated in part (a) using components.

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1-22 Chapter 1 E VALUATE : The vertical component of the 100 N pull is less than the 124 N weight so F " must have an upward component if all three forces balance. Figure 1.75 1.76. I DENTIFY : The four displacements return her to her starting point, so ( ) D = A+ B + C " " " " , where A " , B " and C " are in the three given displacements and D " is the displacement for her return.
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