Explanation The magnitude of the vector forces P and Q are 400 N and 300 N respectively. Refer figure P2.92. Write the equation to find the components of the vector P . P = ( P ) ( − cos 30 ° sin 15 ° i + sin 30 ° j + cos 30 ° cos 15 ° k ) (I) Write the equation to find the components of the vector R . Q = ( Q ) ( cos 50 ° cos 20 ° i + sin 50 ° j − cos 50 ° sin 20 ° k ) (II) Write the equation to find the resultant of the two vectors. R = P + Q (III) Here, R is the resultant of the two vectors P and Q Write the equation to find the magnitude of the vector R . | R | = R x 2 + R y 2 + R z 2 (IV) Write the equation to find the direction of vector R . cos θ n = R n R (V) Here, R n is the component of the vector corresponding to the direction n , R is the magnitude of the vector R , and θ n is the angle corresponding to the direction. Conclusion: Substitute 400 N for P in equation (I) to find P . P = ( P ) ( − cos 30 ° sin 15 ° i + sin 30 ° j + cos 30 ° cos 15 ° k ) = ( 400 N ) ( − cos 30 ° sin 15 ° i + sin 30 ° j + cos 30 ° cos 15 ° k ) = ( − 89.678 N ) i + ( 200 N ) j + ( 334.61 N ) k Substitute 300 N for Q in equation (II) to find Q . Q = ( Q ) ( cos 50 ° cos 20 ° i + sin 50 ° j − cos 50 ° sin 20 ° k ) = ( 400 N ) ( cos 50 ° cos 20 ° i + sin 50 ° j − cos 50 ° sin 20 ° k ) = ( 181

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Chapter 2.4, Problem 2.92P

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Chapter 2 Solutions

VECTOR MECHANIC

Ch. 2.1 - Two forces are applied as shown to a hook....Ch. 2.1 - Two forces are applied as shown to a bracket...Ch. 2.1 - Two forces P and Q are applied as shown at point A...Ch. 2.1 - Two forces P and Q are applied as shown at point A...Ch. 2.1 - A stake is being pulled out of the ground by means...Ch. 2.1 - A telephone cable is clamped at A to the pole AB....Ch. 2.1 - A telephone cable is clamped at A to the pole AB....Ch. 2.1 - A disabled automobile is pulled by means of two...Ch. 2.1 - A disabled automobile is pulled by means of two...Ch. 2.1 - Two forces are applied as shown to a hook support....

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