![In Conceptual Question 4.4: The magnitude of the horizontal force acting on arrow 1 is Select ] the magnitude of the horizontal force acting on arrow 3. The magnitude of the horizontal force acting on arrow 2 is I Select ] the magnitude of the vertical force acting on arrow 3.](http://img.homeworklib.com/questions/17346b10-6fb9-11ea-b8d4-258f30d3c3d6.png?x-oss-process=image/resize,w_560)
Homework Answers
F= (dp/dt)
so all are released in same interval and attain speed at same time .so dt is same .ALL THESE ARE HORIZONTAL FORCES VERTICAL FORCES ARE ZERO AS THERE IS NO VERTICAL ACCELERATION .
F1= 0.080*((10-0)/dt)=0.80/dt
F2= 0.080*((9-0)/dt)=0.72/dt
F3= 0.090*((9-0)/dt)=0.81/dt
HORIZONTAL force acting on arrow 1 F1 is less than horizontal force acting on arrow F3
HORIZONTAL force acting on arrow 2 F2 is greater than vertical force acting on arrow 3 F2>0.
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