Question

1. the largest pumpkin ever grown had a weight of 4.40x10³ n. suppose this pumpkin is placed on a platform that is supported by two bases 5.00 m apart. if the left base exerts an upward normal force of 2.70x10³ n on the platform, how far must the pumpkin be from the platforms left edge? the platform has negligible mass.

Explanation:

Step1: Calculate the upward force exerted by the right - hand base

The total weight of the pumpkin is $W = 4.40\times10^{3}\ N$, and the upward force exerted by the left - hand base is $F_{1}=2.70\times10^{3}\ N$. Let the upward force exerted by the right - hand base be $F_{2}$. According to the equilibrium condition of forces in the vertical direction ($\sum F_y = 0$), we have $W=F_{1}+F_{2}$. So, $F_{2}=W - F_{1}$.
$F_{2}=4.40\times10^{3}-2.70\times10^{3}=1.70\times10^{3}\ N$

Step2: Set up the torque equation

Let the length between the two bases be $L = 5.00\ m$, and the distance of the pumpkin from the left - hand base be $x$. Take the left - hand base as the pivot point. The torque due to the weight of the pumpkin is $\tau_{W}=W\times x$ (clock - wise), and the torque due to the upward force of the right - hand base is $\tau_{F_{2}}=F_{2}\times L$ (counter - clockwise). According to the equilibrium condition of torques ($\sum\tau = 0$), we have $\tau_{W}=\tau_{F_{2}}$.
$W\times x=F_{2}\times L$

Step3: Solve for the distance $x$

We know $W = 4.40\times10^{3}\ N$, $F_{2}=1.70\times10^{3}\ N$, and $L = 5.00\ m$. From $x=\frac{F_{2}\times L}{W}$, we substitute the values:
$x=\frac{1.70\times10^{3}\times5.00}{4.40\times10^{3}}\ m\approx1.93\ m$

Answer:

$1.93\ m$