Chapter 3

Chapter 3 - 3.1. Visualize: Solve: (a) To find A + B , we...

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3.1. Visualize: Solve: (a) To find A B + G G , we place the tail of vector B G on the tip of vector A G and connect the tail of vector A G with the tip of vector . B G (b) Since () A BA B −=+− GG , we place the tail of the vector B G on the tip of vector A G and then connect the tail of vector A G with the tip of vector B G .
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3.2. Visualize: Solve: (a) To find A B + G G , we place the tail of vector B G on the tip of vector A G and then connect vector ’s A G tail with vector ’s B G tip. (b) To find A B G G , we note that () A BA B −=+− GG . We place the tail of vector B G on the tip of vector A G and then connect vector ’s A G tail with the tip of vector . B G
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3.3. Visualize: Solve: Vector E G points to the left and up, so the components x E and y E are negative and positive, respectively, according to the Tactics Box 3.1. (a) cos x EE θ =− and sin . y = (b) sin x φ and cos . y = Assess: Note that the role of sine and cosine are reversed because we are using a different angle. and are complementary angles.
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3.4. Visualize: The position vector r G whose magnitude r is 10 m has an x -component of 6 m. It makes an angle θ with the -axis x + in the first quadrant. Solve: Using trigonometry, cos , x rr = or 6 m (10 m)cos . = This gives 53.1 . = ° Thus the y -component of the position vector r G is sin (10 m)sin53.1 8 m. y == ° = Assess: The y -component is positive since the position vector is in the first quadrant.
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3.5. Visualize: The figure shows the components v x and v y , and the angle θ . Solve: We have, sin 40 , y vv =− ° or 10 m/s sin 40 , v −= −° or 15.56 m/s. v = Thus the x -component is cos40 (15.56 m/s ) cos40 12 m/s. x = ° = Assess: The x -component is positive since the position vector is in the fourth quadrant.
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3.6. Visualize: We will follow rules in Tactics Box 3.1. Solve: (a) Vector r G points to the right and down, so the components x r and y r are positive and negative, respectively: cos (100 m)cos45 70.7 m x rr θ == ° = sin (100 m)sin 45 70.7 m y =− °=− (b) Vector v G points to the right and up, so the components x v and y v are both positive: cos (300 m /s) cos20 282 m/s x vv ° = sin (300 m/s)sin 20 103 m/s y ° = (c) Vector a G has the following components: 22 cos (5.0 m/s )cos90 0 m/s x aa °= sin (5.0 m/s )sin90 5.0 m/s y Assess: The components have same units as the vectors. Note the minus signs we have manually inserted according to Tactics Box 3.1.
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3.7. Visualize: We will follow the rules given in Tactics Box 3.1. Solve: (a) (5 cm/s)sin90 5 cm/s x v =− °=− (5 cm/s)cos90 0 cm/s y v = (b) 22 (10 m/s )sin40 6.4 m/s x a (10 m/s )cos40 7.7 m/s y a (c) (50 N)sin36.9 30 N x F = (50 N)cos36.9 40 N y F = Assess: The components have the same units as the vectors. Note the minus signs we have manually inserted according to Tactics Box 3.1.
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3.8. Visualize: The components of the vector C G and , D G and the angles θ are shown. Solve: For C K we have (3.15 m)cos15 3.04 m x C =− °=− and (3.15 m)sin15 0.815 m.
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This note was uploaded on 11/26/2010 for the course MTH 232 taught by Professor Smith during the Spring '10 term at UConn.

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Chapter 3 - 3.1. Visualize: Solve: (a) To find A + B , we...

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