Kein Zurück: A Newton's Laws of Motion Problem!

Two adults and a child want to push a wheeled cart in the horizontal direction x. The two adults push with horizontal forces $\small F_{1}$ and $\small F_{2}$ as shown in the free-body diagram. (a) Find the magnitude and direction of the smallest force that the child should exert. You can ignore the effects of friction. (b) If the child exerts the minimum force found in part (a), the cart accelerates at $\small 2.0m/s^{2}$ in the +x-direction. What is the weight of the cart?

(a)If the box is pushed in the +x-direction by the total force of the two adults and the child...

$a_{y}=0\;\;\;\;\; so \;\;\;\;\sum F_{y}=0$

$\sum F_{y}=F_{1y}+F_{2y}+F_{3y}=0$

$F_{3y}=-(F_{1y}+F_{2y})$

$F_{1y}=100\sin\frac{\pi}{3}\;\;\;\;F_{2y}=140sin-\frac{\pi}{6}$

$F_{3y}=-[86.6+(-70)]=-16.6=-17N=17N$
$F_{3x}=0$

Because the question asks for the magnitude, take the absolute value of the final answer -17N, which is 17N. The direction in which the little girl is pushing is 90 degrees with respect to the x-axis.

(b) Because we know the x-components of the forces, we can find the mass, which will allow us to calculate the weight of the cart by w=mg.

$F_{1x}=100\cos \frac{\pi}{3}=50N$

$F_{2x}=140\cos \frac{\pi}{6}=121.2N$

$\sum F_{x}=F_{1x}+F_{2x}=50N+121.2N=171.2N$

$m=\frac{\sum F_{x}}{a}=\frac{171.2N}{2.0m/s^{2}}=85.6\;kg$

$w=mg=(85.6kg)(9.8m/s^{2})=840N\therefore$

Kein Zurück

Saturday, June 27, 2009

A Newton's Laws of Motion Problem!

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