Some of the content of this guide was modeled after a guide published by OpenStax and has been adapted for the NWP Learning Commons in March 2022. This work is licensed under a Creative Commons BY 4.0 International License.
The first step is to define an appropriate reference axis. One approach is to make the reference axis parallel and perpendicular to the main direction of motion, which in this case is the slope itself. The next step is to find the parallel and perpendicular forces acting on the block in all 4 main directions. The difference with this example and the one from the previous examples page is the force of kinetic friction that works in the \(x_1-axis\).
First, we need to find the angle of the incline plane:
$$Tan\,\theta=Height/Length={7\,m}/{12\,m}$$
$$\theta=tan^{-1}\,({7\,m}/{12\,m})\,\cong30^o$$
This angle is the same angle that \(F_g\) makes with the \(y_1\) axis.
\[\vec{\textbf{F}}_g=F_{gx_1}\textbf{i}-F_{gy_1}\textbf{j}\]
\[\vec{\textbf{F}}_g=F_{g}*sin\,\theta\,\textbf{i}-F_{g}*cos\,\theta\,\textbf{j}\]
The next step is to check the Net External Force for each dimension of the reference axis:
$$\sum{F_{y1}}=F_{N}-F_{g}*cos\,\theta=0$$
$$\sum{F_{x1}}=F_{gx1}-F_{f_k}=m_ba$$
The net external force in the \(y_1\) axis is zero as the block is not falling through or lifting off of the surface. This equation is required for the normal force, as we introduced friction on the surface of the incline plane.
Now combine the equations and simplify:
$$\sum{F_{y1}}=F_{N}-F_{g}*cos\,\theta=0$$
$$F_{N}=F_{g}*cos\,\theta$$
$$\sum{F_{x1}}=F_{gx1}-F_{f_k}=m_ba$$
$$F_{g}*sin\,\theta-F_{N}*\mu_k=m_ba$$
$$F_{g}*sin\,\theta-F_{g}*cos\,\theta*\mu_k=m_ba$$
$$m_bg*sin\,\theta-m_bg*cos\,\theta*\mu_k=m_ba$$
$$g*sin\,\theta-g*cos\,\theta*\mu_k=a$$
$$a=g*(sin\,\theta-\mu_kcos\,\theta)$$
$$a=9.81m/s^2*(sin30-(0.25)*cos30)$$
$$a=2.78m/s^2$$
Identify and separate the bodies involved in the problem. Assign appropriate internal forces acting on each body.
Body 1:
$$\sum{F_{y}}=0$$
$$T-W=0,\;T=W=m_1g$$
Body 2:
$$T=T$$
Body 3:
$$\sum{F_{y}}=0$$
$$F_N-W=0,\;F_N=W=m_2g$$
$$\sum{F_{x}}=0$$
$$T-F_F=0,\;F_F=T$$
As the tension in the rope is equivalent to the weight of Body 1, the force of friction can be calculated as:
$$F_F=T=m_1g=25kg*9.81m/s^2=245N$$
If this force of friction is the upper limit of the static friction, we can use the normal force of the second block to calculate the coefficient of static friction:
$$F_F=\mu_sF_N$$
$$\mu_s=\frac{F_F}{F_N}=\frac{F_F}{m_2g}$$
$$\mu_s=\frac{245N}{35kg*9.81m/s^2}=0.714$$