#### Answer

(a) The child should exert a force of 16.6 N in the direction of the (-y)-axis.
(b) The weight of the cart is 840 N.

#### Work Step by Step

(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.