# Question #5a331

Feb 26, 2017

The acceleration of the refrigerator is $0.49 \frac{m}{s} ^ 2$ when the minimum force is applied.

#### Explanation:

We are given that ${\mu}_{k} = 0.20 , {\mu}_{s} = 0.25$, and $m = 100.0 k g$. In order to determine the acceleration of the refrigerator when the minimum force is applied, we first need to determine the minimum force, i.e. that which is necessary to overcome the force of static friction.

Where $\vec{n}$ is the normal force, ${\vec{F}}_{g}$ is the force of gravity, ${\vec{f}}_{s}$ is the force of static friction, and ${\vec{F}}_{p}$ is the force of the push applied to the refrigerator. I have defined to the right as the positive direction, i.e. all force vectors pointing to the right are positive and those pointing to the left are negative. As usual, force vectors pointing up are positive, and those pointing down are negative.

(Note that not all of the vector magnitudes are necessarily drawn to scale)

We can construct statements for the net force in both the parallel $\left(x\right)$ and perpendicular $\left(y\right)$ direction.

$\sum {\vec{F}}_{x} = {\vec{F}}_{p} - {\vec{f}}_{s}$

$\sum {\vec{F}}_{y} = \vec{n} - {\vec{F}}_{g}$

Because the refrigerator does not move in the perpendicular direction (up or down), the net force ${\vec{F}}_{y}$ must be zero. Recall that force is given as $\vec{F} = m \vec{a}$ by Newton's second law. We can see from this equation that if there is no force acting on an object, it can have no acceleration (mass cannot be zero). We want the minimum force required to get the refrigerator to move, so we are looking for a horizontal acceleration.

$\implies \sum {\vec{F}}_{x} = {\vec{F}}_{p} - {\vec{f}}_{s} = m {\vec{a}}_{x}$

$\implies \sum {\vec{F}}_{y} = \vec{n} - {\vec{F}}_{g} = 0$

We know that the equation for the maximum static friction is given by ${\vec{f}}_{s} = {\mu}_{s} \vec{n}$.

In order for the refrigerator to move, or to accelerate, we must overcome the force of static friction. Therefore, we will calculate the maximum force of static friction.

From our statement of the net force, we see that $\vec{n} = {\vec{F}}_{g}$, and as ${\vec{F}}_{g} = m g$, we have $\vec{n} = m g$.

$\implies {\vec{f}}_{s} = {\mu}_{s} \cdot m g$

$= \left(0.25\right) \left(100.0 k g\right) \left(9.8 \frac{m}{s} ^ 2\right)$

$= 245 N$.

We must apply a force$\ge 245 N$.

Now with a value for the force, we can calculate the acceleration of the refrigerator. Once we overcome static friction, we have:

This is very similar to what we had above, except now we have the force of kinetic friction, as the object has begun to move.

$\implies \sum {\vec{F}}_{x} = {\vec{F}}_{p} - {\vec{f}}_{k} = m {\vec{a}}_{x}$

$\implies \sum {\vec{F}}_{y} = \vec{n} - {\vec{F}}_{g} = 0$

$\implies {\vec{f}}_{k} = {\mu}_{k} \vec{n} = {\mu}_{k} \cdot m g$

Now we want to calculate $\vec{a}$. We will use ${\vec{F}}_{p} - {\vec{f}}_{k} = m {\vec{a}}_{x}$ from above.

$\implies {\vec{F}}_{p} - \left({\mu}_{k} \cdot m g\right) = m {\vec{a}}_{x}$

Solve for $\vec{a}$:

$\implies \frac{{\vec{F}}_{p} - \left({\mu}_{k} \cdot m g\right)}{m} = {\vec{a}}_{x}$

Using our given values and what we calculated for ${\vec{F}}_{p}$:

${\vec{a}}_{x} = \frac{245 N - \left(0.20 \cdot \left(100.0 k g\right) \left(9.8 \frac{m}{s} ^ 2\right)\right)}{100.0 k g}$

$\implies \frac{49 k g \frac{m}{s} ^ 2}{100.0 k g}$

$\implies 0.49 \frac{m}{s} ^ 2$

$\therefore$ The acceleration of the refrigerator is $0.49 \frac{m}{s} ^ 2$.