Question #31f43

1 Answer
Jane
Jan 16, 2018

#6.874 m/s^2# about 71.90 degrees w.r,t north

Explanation:

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#15 N# force is acting along south and #50N# along northeast. So total force acting along north is #50cos 45# or,35.35 N,so net force acting along north is #(35.35-14)# or,#21.35N#

And force acting along east is #(50 cos 45+30)# or,#65.35N#.

so,let ,net force acting along #theta# degrees w.r,t North and its magnitude is #sqrt##(21.35^2+65.35^2) N# or,#68.74 N#

#tan theta = 65.35/21.35 # or,# theta= 71.90# degrees

so,magnitude of acceleration will be #68.74/10 # or # 6.874m/s^2# along the direction of net force.

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