For Future Reference - Forces as Vectors

Step 1: Set up your Free Body Diagram

This is a crucial step in order to understand what’s going on in the given system. We know that the cart is moving towards the right, but in order to determine the acceleration, we must take into account all the forces acting on the object, which in this case is the cart (view next image in the gallery for FBD).

Step 2: Use FBD to determine all x and y components of forces

Now, you can use the FBD created in step 1 to determine all your x and y components of forces. Notice in the FBD that the friction, gravitational, and normal forces all lie parallel with the reference x and y axes. The applied “pulling” force by the man is done at an angle, so we must break the Force vector into it’s x and y components.

As shown in this gallery, the only two forces acting in the x direction are the friction force (${\mu }_k{F}_N$) and the x-component of the applied force. In the y-direction, there are three total forces. Both normal force and y-component of the applied force in the positive y-direction, and gravitational force ($m\overrightarrow{g}$) acting in negative y-direction.

Step 3: Set up equations of motion

Next is to set up the equations of motion using the force equations for both the x and y components.

$\mathrm{\Sigma }{\overrightarrow{F}}_x=m{\overrightarrow{a}}_x$​
$\mathrm{\Sigma }{\overrightarrow{F}}_y=m{\overrightarrow{a}}_y$​
Using the third image we developed in the gallery and the approach we used in step 2, our equations of motion are:

X component: $9.06\text{N}-{\mu }_k{\overrightarrow{F}}_N=m{\overrightarrow{a}}_x$

Y component, using positive scalar magnitudes: 4.23 N + N − mg = maᵧ, with g = 9.81 m/s².

x-component of applied force :

${\overrightarrow{F}}_{app,x}=10\cdot \cos {25}^{\circ }=9.06\text{N}$​
y-component of applied force:

${\overrightarrow{F}}_{app,y}=10\cdot \sin {25}^{\circ }=4.23\text{N}$​
${\overrightarrow{F}}_g=m\overrightarrow{g}=3\text{kg}\cdot -9.81\text{m}/{\text{s}}^2$

Notice here that we are subtracting the gravitational force $m\overrightarrow{g}$. This is because the gravity force vector is pointing in the negative y-direction. The same with the friction force. It is being subtracted from the x-component of the applied force because it is pointed in the negative x-direction.

Step 4: Find the unknowns

Our first unknown is the normal force, ${\overrightarrow{F}}_N$ .

In order to solve for ${\overrightarrow{F}}_N$, we must use our equations of motion that we prepared in step 3. We cannot use the x-component of the force, because there are two unknowns in the problem, ${\overrightarrow{F}}_N$ and ${\overrightarrow{a}}_x$. We don’t know the acceleration in the x-direction.

The cart stays in contact with the horizontal surface and maintains a fixed height, so its vertical velocity remains zero and aᵧ = 0. Use the y-force balance to solve for N.

Vertical balance: 10 sin 25° + N − 3(9.81) = 0. Thus N = 29.43 − 4.22618 = 25.2038 N ≈ 25.20 N.

Step 5: Solve for the cart’s acceleration

As stated in step 4, there is no acceleration in the y-direction, so that means the cart’s acceleration is all in the x-direction. Now that we’ve solved for the normal force ${\overrightarrow{F}}_N$, and we know ${\mu }_k=0.3$ and $m=3\text{kg}$, we can now solve for ${\overrightarrow{a}}_x$ using the $\mathrm{\Sigma }{\overrightarrow{F}}_x$ equation:

Horizontal balance: 10 cos 25° − 0.3N = 3aₓ. Substituting N = 25.2038 N gives 9.06308 − 7.56115 = 3aₓ.

aₓ = 0.500644 m/s² ≈ 0.501 m/s² to the right.

In the y-direction, there is the normal force ${\overrightarrow{F}}_N$, which is perpendicular to the plane of contact. In the x-direction, there is the frictional force ${\overrightarrow{F}}_{fr}$ , which is pointed in the negative x-direction.

Step 2: Use FBD to determine all x and y components of forces

Now that the FBD and orientation are established, we can analyze the system to separate the forces into their respective x and y components. See the gallery here to view the component forces for each direction, x and y. ​
Notice how the gravitational force was broken up into two separate components. The components are done to follow the reference axes, so the x and y components of the gravitational force are, respectively:

${\overrightarrow{F}}_{g,x}=mg\cdot \sin {40}^{\circ }=10\text{kg}\cdot 9.81\frac{\text{m}}{{\text{s}}^2}\cdot \sin {40}^{\circ }=63.1\text{N}$

${\overrightarrow{F}}_{g,y}=mg\cdot \cos {40}^{\circ }=10\text{kg}\cdot 9.81\frac{\text{m}}{{\text{s}}^2}\cdot \cos {40}^{\circ }=75.2\text{N}$

The image also shows two other forces acting on the box.

Step 4: Find the unknown normal force, ${\overrightarrow{F}}_N$

Now, we need to solve for ${\overrightarrow{F}}_N$ in order to find our coefficient of static friction. How exactly are we going to do this? Well, remember in the problem statement that the box is at rest on the inclined plane. It being at rest means that it is not moving (i.e “static”).

Therefore, the box is neither accelerating in the x or y direction. Meaning then that ${\overrightarrow{a}}_y$ and ${\overrightarrow{a}}_x$ values are equal to ZERO.

Now our equations are:

$63.1\text{N}-{\mu }_s{\overrightarrow{F}}_N=m\cdot 0$ ​
${\overrightarrow{F}}_N-75.2\text{N}=m\cdot 0$

Solving for ${\overrightarrow{F}}_N$ gives us a value of ${\overrightarrow{F}}_N=75.2\text{ N}$.

Step 1: Set up your Free Body Diagram

Just like the previous example problem, it is imperative that an FBD is created. Look at this problem’s gallery’s second image to see the FBD. Something new to notice in the FBD is that there is no applied force acting on the box. Also notice how my reference x and y-axes are oriented differently than before. We shifted the axes’ orientations to match the orientation of the box.

Changing the orientation to that of the box’s makes the problem much easier to understand visually and it can be done for any problem where the object you are trying to analyze is in a crazy orientation!

Step 3: Set up your equations of motion (EOMs)

Let’s set up our EOMs for both the x and y components. Again we will be using both components of the force equations:

$\mathrm{\Sigma }{\overrightarrow{F}}_x=m{\overrightarrow{a}}_x$​
$\mathrm{\Sigma }{\overrightarrow{F}}_y=m{\overrightarrow{a}}_y$​
Now, using the third image in the gallery and the equations above, let’s plug in the forces for both the x and y components.

x - component:

${\overrightarrow{F}}_{g,x}-{\overrightarrow{F}}_{fr}=m{\overrightarrow{a}}_x$ ​
$63.1\text{N}-{\mu }_s{\overrightarrow{F}}_N=m{\overrightarrow{a}}_x$

y - component:

${\overrightarrow{F}}_N-{\overrightarrow{F}}_{g,y}=m{\overrightarrow{a}}_y$ ​
${\overrightarrow{F}}_N-75.2\text{N}=m{\overrightarrow{a}}_y$

Step 5: Solve for coefficient of static friction

At the threshold of sliding, fₛ = μₛ,minN. With N = mg cos 40° and the required friction mg sin 40°, μₛ,min = tan 40° ≈ 0.839. Any larger coefficient can also hold the box at rest.

${\mu }_s=0.84$

Now that is a pretty high ${\mathit{\mu }}_{\mathit{s}}$ value!

Section breakdown

1. Friction
2. Force Vectors
3. Example 1.4.1 - The Man who Struck Gold
4. Example 1.4.2 - The Precariously Placed Box
5. Chapter Equations
6. Chapter Problems