## University Physics with Modern Physics (14th Edition)

(a) We can find the y-component of the sum of the forces of the two adults. $\sum F_y = F_{1y}+F_{2y}$ $\sum F_y = (100~N)~sin(60^{\circ}) - (140~N)~sin(30^{\circ})$ $\sum F_y = 16.6~N$ (+y direction) To use the minimum possible force, the child should exert force only in the direction of the (-y)-axis to oppose the sum of the adults' forces in the direction of the +y-axis. Therefore, the child should exert a force of 16.6 N in the direction of the (-y)-axis. (b) We can use a force equation to find the mass of the cart. $\sum F_x = ma$ $F_{1x}+F_{2x} = ma$ $m = \frac{(100~N)~cos(60^{\circ}) + (140~N)~cos(30^{\circ})}{2.0~m/s^2}$ $m = 85.6~kg$ We can use the mass to find the weight of the cart. $weight = mg = (85.6~kg)(9.80~m/s^2)$ $weight = 840~N$ The weight of the cart is 840 N.