Confused about friction forces

Question

"A bedroom bureau with a mass of $45$ kg, including drawers and clothing, rests on the floor. (a) If the coefficient of static friction between the bureau and the floor is $0.45$, what is the magnitude of the minimum horizontal force that a person must apply to start the bureau moving? (b) If the drawers and clothing, with $17$ kg mass, are removed before the bureau is pushed, what is the new minimum magnitude?"

The book explains it with this: https://i.stack.imgur.com/T5Chc.png

I don't understand the explanation for the formulas. Am I supposed to sketch the graph first? And then what? How would I go from the graph to $F - F_{s, max} = ma$ and $F_n - mg = 0$? How are those formulas reached by looking at the graph?

Also, I don't understand how the two equations somehow become $F - \mu_s \cdot mg = ma = 0$. How does that happen?

Answer 1

So, essentially for these questions, always start by drawing your force diagram (it is not really a graph, more of a way to keep track of what all is going on). Then what they are doing is summing the forces in the x-direction to get $F - f_s$ (where $F$ is the force we want to find and $f_s$ is the force of static friction) which is equal to $ma_x$ by Newton's Second Law. Then they sum the forces in the y-direction to get $F_N - mg$ where $F_N$ is the normal force and $mg$ is simply the force of gravity. In this direction, however, there will not be any acceleration. Now, recall that the $f_s = \mu _s F_N$ and $F_N$ must equal $mg$, from the second equation. Now the math is out of the way and we have to think conceptually: we want the minimum force needed, so we only want to barely move the furniture, so we set $a_x = 0$ and solve.

Answer 2

It is better to sketch a graph to understand what the problem says, and the explanation for the what formulas wants to express physical states.

The free-body diagram says about forces which acts on a body. These forces are divided to two components. This is a mathematical explanation for physical states. Here we have forces in $x$ direction and $y$ direction also. All forces obey Newton's laws. So with horizontal forces we have $$F - F_s = ma_x$$ and for vertical forces we have $$F_N - mg = ma_y$$ The body doesn't move in $y$ direction so it has not acceleration in this type, then $a_y=0$.

Answer 3

One should imagine solving this problem on a planar graph.

The force $\vec{F}$ is strictly a horizontal force (as indicated by the word problem), so one could write this as $\vec{F}= \langle |\vec{F}|,0\rangle$. So $\vec{F}$ does not exert a vertical force onto the floor.

The force of static friction is said to be pointing in the negative $x$-direction. So $-\vec{f_s}=\langle -|\vec{f_s}|,0\rangle$.

The normal force $\vec{F_N}$ is the vertical force exerted by the floor. So $\vec{F_N}=\langle 0,|\vec{F_N}|\rangle$.

Finally, you have the gravitational force $m(-\vec{g})$ acting on the object, i.e., it is $\langle 0, -m |\vec{g}|\rangle$ (where without loss of generality, one could think of $\vec{g}$ as $\vec{g}=\langle 0, 9.8 \mbox{ m/s}^2\rangle$). The minus sign appears since the gravity is pointing downwards.

If the object were stable (non-moving), then we add these four vectors to get the resultant force vector $\langle 0,0\rangle$.

That is, $$ \vec{F} - \vec{f_s} + \vec{F_N} - m\vec{g} = \langle |\vec{F}|,0\rangle + \langle -|\vec{f_s}|,0\rangle + \langle 0,|\vec{F_N}|\rangle + \langle 0, -m |\vec{g}|\rangle =\langle 0,0\rangle. $$

Since we add vectors component-wise, we see that $$ |\vec{F}| -|\vec{f_s}|+0+0 = 0 \mbox{ and } 0+0+ |\vec{F_N}| -m |\vec{g}|=0. $$ So $$ |\vec{F}| -|\vec{f_s}| = 0 \mbox{ and } |\vec{F_N}| -m |\vec{g}|=0. $$ Since the $y$-components of $\vec{F}$ and $\vec{f_s}$ add to zero, we have $\vec{F} - \vec{f_s} = \langle 0,0\rangle$, and since the $x$-components of $ \vec{F_N}$ and $m\vec{g}$ add to zero, we have $\vec{F_N} - m\vec{g} = \langle 0,0\rangle$.

Thus, the two equations \begin{align*} \vec{F} - \vec{f_s} &= \langle 0,0\rangle \\ \vec{F_N} - m\vec{g} &= \langle 0,0\rangle \\ \end{align*} are precisely the two equations derived in the png attachment.

Now, notice that static friction is denoted by $-\vec{f}_{s,\max}$, and in the case when the object is moving, i.e., it has a nonzero displacement vector (in the $x$-direction), then that is when we have the two equations: \begin{align*} \vec{F} - \vec{f}_{s,\max} &= m\vec{a}= \langle m|\vec{a}|,0 \rangle, \\ \vec{F_N} - m\vec{g} &= \langle 0,0\rangle, \\ \end{align*} where $\vec{a}$ is the resultant acceleration of the object in the positive $x$-direction.

In the case when $-\vec{f}_{s,\max}$ is the $\textbf{maximum}$ static friction $-\vec{f}_{s,\max}=\langle -\mu_s m|\vec{g}| ,0\rangle$, then this would mean that the object is not moving, i.e., the acceleration vector $\vec{a}$ is the zero vector $\langle 0,0\rangle$.