Question

A big olive (* - 0.50 kg) lies at the origin of an xy coordinate system, and a big BrazlI nut (M - 1.5^kg) lie^s at the point (1 .0,2.0) m. At t : 0, a force Fo : Q.Oi 3.0j ) N begins to act on the olive, and a force Fn: (-3.0i - 2.0i) N begins to act on the nut. In unit-vector notation, what is the displacement of the center of mass of the olive

Answer 1

The estimated displacement of the center of mass of the olive is $\overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m]$.

Procedure - Estimation of the displacement of the center of mass of the olive

In this question we should apply the definition of center of mass and difference between the coordinates for dynamic ($\vec r$) and static conditions ($\vec r_{o}$) to estimate the displacement of the center of mass of the olive ($\overrightarrow{\Delta r}$):

$\vec r - \vec r_{o} = \left[\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot(m_{i}\cdot g + F_{i, x})}{\Sigma \limits_{i =1}^{2}(F_{i,x}+m_{i}\cdot g)} ,\frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot(m_{i}\cdot g + F_{i, y})}{\Sigma \limits_{i =1}^{2}(F_{i,y}+m_{i}\cdot g)} \right]-\left(\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}, \frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}\right)$ (1)

Where:

• $r_{i, x}$ - x-Coordinate of the i-th element of the system, in meters.
• $r_{i,y}$ - y-Coordinate of the i-th element of the system, in meters.
• $F_{i,x}$ - x-Component of the net force applied on the i-th element, in newtons.
• $F_{i,y}$ - y-Component of the net force applied on the i-th element, in newtons.
• $m_{i}$ - Mass of the i-th element, in kilograms.
• $g$ - Gravitational acceleration, in meters per square second.

If we know that $\vec r_{1} = (0, 0)\,[m]$, $\vec r_{2} = (1, 2)\,[m]$, $\vec F_{1} = (0, 3)\,[N]$, $\vec F_{2} = (-3, -2)\,[N]$, $m_{1} = 0.50\,kg$, $m_{2} = 1.50\,kg$ and $g = 9.807\,\frac{kg}{s^{2}}$, then the displacement of the center of mass of the olive is:

Dynamic condition$\vec{r} = \left[\frac{(0)\cdot (0.50)\cdot (9.807)+(0)\cdot (0) + (1)\cdot (1.50)\cdot (9.807) + (1)\cdot (-3)}{(0.50)\cdot (9.807) + 0 + (1.50)\cdot (9.807)+(-3)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (0)\cdot (3) + (2)\cdot (1.50)\cdot (9.807) +(2) \cdot (-2)}{(0.50)\cdot (9.807) + (3)+(1.50)\cdot (9.807)+(-2)} \right]$$\vec r = (0,704, 1.233)\,[m]$

Static condition

$\vec{r}_{o} = \left[\frac{(0)\cdot (0.50)\cdot (9.807) + (1)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807) + (1.50)\cdot (9.807)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (2)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807)+(1.50)\cdot (9.807)} \right]$

$\vec r_{o} = \left(0.75, 1.50)\,[m]$

Displacement of the center of mass of the olive

$\overrightarrow{\Delta r} = \vec r - \vec r_{o}$

$\overrightarrow{\Delta r} = (0.704-0.75, 1.233-1.50)\,[m]$

$\overrightarrow{\Delta r} = (-0.046, -0.267)\,[m]$

The estimated displacement of the center of mass of the olive is $\overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m]$. $\blacksquare$

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