The free-body diagram of the given situation is shown below:
Fr is the friction force, N is the normal force, Mgcosθ and Mgsinθ are the components of the weight for the case of a body on an incline.
The friction force is given by:
![F_r=\mu N](https://img.qammunity.org/2023/formulas/physics/college/f3rxu92htq7aykq8lwpwygxfs8h5vqim6f.png)
Due to the system is in equilibrium for the forces perpendicular to the plane, you have:
![\begin{gathered} N-Mg\cos \theta=0 \\ N=Mg\cos \theta \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/je71bqnr65jqp980szeg09p0cqlvvn0sgt.png)
Then, for Fr:
![F_r=\mu Mg\cos \theta](https://img.qammunity.org/2023/formulas/physics/college/fb7c50k0wy8ob1r8d9gj9ye8ic812gstnp.png)
where,
θ: angle of the incline = 40
M: mass of the body = 10 kg
g: gravitational acceleration constant = 9.8 m/s^2
Replac