Answer to Question #201725 in Mechanics | Relativity for Hussain Darboe

Question

Two adults and a child want to push a wheeled cart in the +x direction. The two adults push with horizontal forces F1=100N at 60 degrees from +x anticlockwise and F2+140N at 30 degrees from the +x clockwise. (a) Find the magnitude and the direction of the small force that the child should exert, ignoring friction. (b) If the child exerts the minimum force in part (a), the cart accelerates at 2m/s^2 in the +x direction. What is the weight of the cart?

Answer 1

a)

Forces in the +y-direction:

$F_{1y}=100sin60\degree=100\sqrt{3}/2=86.6\ N$

$F_{2y}=-140sin30\degree=-140/2=-70\ N$

So, the smallest force that the child should exert:

$F=86.6-70=16.6\ N$ at 90$\degree$ from +x clockwise (-y-direction).

b)

Result forces in the +x-direction:

$F_x=F_{1x}+F_{2x}=100cos60\degree+140cos30\degree=100/2+140\sqrt{3}/2=171.24\ N$

The weight of the cart:

$m=F_{x}/a=171.24/2=85.62\ kg$