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File Edit View History Bookmarks People Window Help me - myLSU ecure https://edugen.wileyplus.com/edugen/student/mainfr.uni Mail Tools K Mail-jbra45@lsu.edu xCourse: 2017 Fall CM 2113 x WillyPLUS n Assignment Chapter 04, Problem 011 Only two forces act on an object (mass 3.19 kg), as in the drawing. Find (a) the magnitude and (b) the direction (relative to the x axis) of the scceleration of the object m 13 45.0 60.0 N 40.0 m 22 m 22 (a) Number m.35 (b) Number Units Units Question Attempts: Unlimited SAVE FOR LATERSURMET ANSWER Version 4.24.1.20 com/edugen/shared/assignment/test/aglist unid Inc. All Rights Reserved. A Division of 3ohn Wiley & Sons, Inc 8

Explanation / Answer

Given

forces acting on the mass m = 3.19 kg are

let

F1 = 40 N along the +ve x axis and

F2 = 60 N making 45 degrees with +v x axis (or with F1)

the angle between the vectors is 45 degrees

now writing the components of the forces

F1 = F1x i + F1y j = F1 cos 0 i + F1 sin0 j = 40 N i + 0 N j

F2 = F2x i + F2y j = F2 cos 45 i + F2 sin45 j = 60cos45 N i + 60sin45 N j

F2 = 42.42641 N i + 42.42641 N j

the net force is F = F1+F2 = (F1x+F2x)i +(F1y+F2y)j

F = (40+42.42641)N i + (0+42.42641 )N j

F = (82.42641) N i + (42.42641) N j

the magnitude is

F = sqrt((82.42641)^2+(42.42641)^2) N

F = 92.70 N

we know that the force F = ms ==> a = F/m = 92.70/3.19 m/s2 = 29.1 m/s2

the direction is theta = arc tan (Fy/Fx) = arc tan(42.42641/82.42641) s = 27.24 degrees

the direction of the acceleration of the mass is 27.24 degrees relative to the x axis

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