Explanation Given information: The angle between two 150 N forces is 50 ° and the magnitude of the resultant force is 600 N . The below figure represent the forces acting at point A . Figure-(1) Write the expression of the magnitude of the resultant force. R = R x 2 + R y 2 …… (I) Here, the total force applied on the x direction is R x and the total force applied on y direction is R y . Write the expression of total force applied on x direction by using Figure-(1) R x = ( F 1 ) x + ( F 2 ) x + ( F 3 ) x …… (II) Here, the x component of first force is ( F 1 ) x , the x component of second force is ( F 2 ) x and the x component of third force is ( F 3 ) x . Write the expression of total force applied on y direction by using Figure-(1) R y = ( F 1 ) y + ( F 2 ) y + ( F 3 ) y …… (III) Here, the y component of first force is ( F 1 ) y , the y component of second force is ( F 2 ) y and the y component of third force is ( F 3 ) y . Write the expression of the x component of first force. ( F 1 ) x = − F 1 cos θ 1 Here, the first force is F 1 and the angle between first force and y axis is θ 1 . Write the expression of the y component of first force. ( F 1 ) y = F 1 sin θ 1 Write the expression of the x component of second force. ( F 2 ) x = − F 2 cos α Here, the second force is F 2 and the angle between first force and y axis is α . Write the expression of the y component of second force. ( F 2 ) y = − F 2 sin α Write the expression of the x component of third force. ( F 3 ) x = − F 3 cos ( α + θ 2 ) Here, the third force is F 3 and the angle between third force and second force is θ 2 . Write the expression of the y component of second force. ( F 3 ) y = − F 3 sin ( α + θ 2 ) Substitute − F 1 cos θ 1 for ( F 1 ) x , − F 2 cos α for ( F 2 ) x and − F 3 cos ( α + θ 2 ) for ( F 3 ) x in Equation (II) R x = − F 1 cos θ 1 − F 2 cos α − F 3 cos ( α + θ 2 ) …… (IV) Substitute F 1 sin θ 1 for ( F 1 ) y , − F 2 sin α for ( F 2 ) y and − F 3 sin ( α + θ 2 ) for ( F 3 ) y in Equation (III) R y = F 1 sin θ 1 − F 2 sin α − F 3 sin ( α + θ 2 ) …… (V) Calculation: Substitute 500 N for F 1 , 30 ° for θ 1 , 150 N for F 2 , 150 N for F 3 and 50 ° for θ 2 in Equation (IV) R x = − ( 500 N ) cos ( 30 ° ) − ( 150 N ) cos α − ( 150 N ) cos ( α + 50 ° ) = − ( 433.01 N ) − ( 150 N ) cos α − ( 150 N ) { ( cos α ) ( cos 50 ° ) − ( sin α ) ( sin 50 ° ) } = − ( 433.01 N ) − ( 150 N ) cos α − ( 96.42 N ) cos α + ( 114.91 N ) sin α = − ( 433.01 N ) − ( 246.42 N ) cos α + ( 114.91 N ) sin α Substitute 500 N for F 1 , 30 ° for θ 1 , 150 N for F 2 , 150 N for F 3 and 50 ° for θ 2 in Equation (V) R y = ( 500 N ) sin ( 30 ° ) − ( 150 N ) sin α − ( 150 N ) sin ( α + 50 ° ) = ( 250 N ) − ( 150 N ) sin α − ( 150 N ) { ( sin α ) ( cos 50 ° ) + ( sin 50 ° ) ( cos α ) } = ( 250 N ) − ( 150 N ) sin α − ( 96

2nd Edition

ISBN: 9780073398167

Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David Mazurek

Publisher: McGraw-Hill Education

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Chapter 2, Problem 110RP

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The range of the value of angle α.

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The range of the value of angle α .

Solution Summary: The author explains the range of the value of angle alpha , which is 56.98°, and the magnitude of resultant force.

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