Two adults and a child want to push a wheeled cart in the direction marked X ( a
ID: 1915359 • Letter: T
Question
Two adults and a child want to push a wheeled cart in the direction marked X ( axis x positive). The two adults push with horizontal force F1 and F2 (one in each corner of the cart in same side). The angle to positive quadrant I two axis x is 60 degree and the other angle in quadrant IV to axis x is 30 degree. (F1= 100N in the 60 degree angle and F2=140N in the 30 degree angle) The child is in the positive y axis . a) Find the magnitude and direction of the smallest force that the child should exert. you can ignore the effect of friction. b) If the child exert the minimum force found in part (a), the cart accelerates at 2.0m/s^2 in the + x-direction. What is the weight of the cart?Explanation / Answer
this is the method it appears that the child puts in enough force to lift the skids of the cart off the ground Adult 1 is pushing DOWN @ 60 degrees 100N The vertical component of that push is sin 60 = Y1/100 Y1 = 86.6 N DOWN ===> - 86.4 N Adult 2 is pushing UP @ 30 degrees 140 N Vertical component is sin 30 = Y2/140 Y2 = 70 N UP ===> +70 Net Force by by the adults is 70 - 86.6 = - 16.6 N ===> Child must put in 16.6 N UP to lift the skids so the cart will move (a) Now we need the horizontal components of the adults, the child has no horizontal component. cos 60 = X1/100 X1 = 50 N cos 30 = X2/140 X2 = 121.24 N Total: 121.24 + 50 = 171.24 N (Note the X components are both in the + X direction That component lies exactly on the X axis not 5.54 degrees above. Re-check your calculation and think it through again) F = MA 171.24 N = M * 2 m/s^2 M = 171.24/2 = 85.62 kg But we need weight F = MA = 85.62 * 9.8 m/s^2 = 839.09 N