We are given a box that slides up a ramp. To determine the force of friction we will use the following relationship:
![F_f=\mu N](https://tex.z-dn.net/?f=F_f%3D%5Cmu%20N)
Where.
![\begin{gathered} N=\text{ normal force} \\ \mu=\text{ coefficient of friction} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20N%3D%5Ctext%7B%20normal%20force%7D%20%5C%5C%20%5Cmu%3D%5Ctext%7B%20coefficient%20of%20friction%7D%20%5Cend%7Bgathered%7D)
To determine the Normal force we will add the forces in the direction perpendicular to the ramp, we will call this direction the y-direction as shown in the following diagram:
In the diagram we have:
![\begin{gathered} m=\text{ mass}_{} \\ g=\text{ acceleration of gravity} \\ mg=\text{ weight} \\ mg_y=y-\text{component of the weight. } \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20m%3D%5Ctext%7B%20mass%7D_%7B%7D%20%5C%5C%20g%3D%5Ctext%7B%20acceleration%20of%20gravity%7D%20%5C%5C%20mg%3D%5Ctext%7B%20weight%7D%20%5C%5C%20mg_y%3Dy-%5Ctext%7Bcomponent%20of%20the%20weight.%20%7D%20%5Cend%7Bgathered%7D)
Adding the forces in the y-direction we get:
![\Sigma F_y=N-mg_y](https://tex.z-dn.net/?f=%5CSigma%20F_y%3DN-mg_y)
Since there is no movement in the y-direction the sum of forces must be equal to zero:
![N-mg_y=0](https://tex.z-dn.net/?f=N-mg_y%3D0)
Now we solve for the normal force:
![N=mg_y](https://tex.z-dn.net/?f=N%3Dmg_y)
To determine the y-component of the weight we will use the trigonometric function cosine:
![\cos 40=\frac{mg_y}{mg}](https://tex.z-dn.net/?f=%5Ccos%2040%3D%5Cfrac%7Bmg_y%7D%7Bmg%7D)
Now we multiply both sides by "mg":
![mg\cos 40=mg_y](https://tex.z-dn.net/?f=mg%5Ccos%2040%3Dmg_y)
Now we substitute this value in the expression for the normal force:
![N=mg\cos 40](https://tex.z-dn.net/?f=N%3Dmg%5Ccos%2040)
Now we substitute this in the expression for the friction force:
![F_f=\mu mg\cos 40](https://tex.z-dn.net/?f=F_f%3D%5Cmu%20mg%5Ccos%2040)
Now we substitute the given values:
![F_f=(0.2)(10\operatorname{kg})(9.8\frac{m}{s^2})\cos 40](https://tex.z-dn.net/?f=F_f%3D%280.2%29%2810%5Coperatorname%7Bkg%7D%29%289.8%5Cfrac%7Bm%7D%7Bs%5E2%7D%29%5Ccos%2040)
Solving the operations:
![F_f=15.01N](https://tex.z-dn.net/?f=F_f%3D15.01N)
Therefore, the force of friction is 15.01 Newtons.