 # Two adults and a child want to push a wheeled cart in thedirection marked x in Fig. 4.29 The two adults push with horizontalforces F_{1} and F_{2} as BenoguigoliB 2021-02-05 Answered

Two adults and a child want to push a wheeled cart in thedirection marked x in Fig. 4.29 The two adults push with horizontalforces  as shown in the figure. (a)Find the magnitude and direction of the smallest force that thechild should exerts the minimum force found in part (a), the cartaccelerates at 2.0 $\frac{m}{{s}^{2}}$ in the +x direction. What is theweight of the cart?

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a) The resultant must have no y-component, and sothe child must push with a force withy-component $140\mathrm{sin}\left(30\right)-100\mathrm{sin}\left(60\right)=16.6N$. For the child to exert the smallestpossible force, that force will have no x-component, so thesmallest possible force has magnitude 16.6 N and is at an angle of 270, or 90 clockwise from the +xdirection.
b) $\sum F=ma$
$m=\frac{\sum F}{a}=\frac{140\mathrm{sin}\left(30\right)+100\mathrm{sin}\left(60\right)}{2}=85.6kg$
$w=mg=\left(85.6\right)\left(9.8\right)=840N$

We have step-by-step solutions for your answer! Jeffrey Jordon

a)Forces in the +y-direction:

${F}_{1y}=100\mathrm{sin}{60}^{\circ }=100\frac{\sqrt{3}}{2}=86.6N$

${F}_{2y}=-140\mathrm{sin}{30}^{\circ }=-\frac{140}{2}=-70N$

So, the smallest force that the child should exert:

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

b)Result forces in the +x-direction:

${F}_{x}={F}_{1x}+{F}_{2x}=100\mathrm{cos}{60}^{\circ }+140\mathrm{cos}{30}^{\circ }=\frac{100}{2}+\frac{140\sqrt{3}}{2}=171.24N$

The weight of the cart:

$m=\frac{{F}_{x}}{a}=\frac{171.24}{2}=85.62kg$

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