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.
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:
$$\sum{F_{y1}}=F_{Ny_{1}}-F_{g}*cos\,\theta=0$$ $$\sum{F_{x1}}=F_{g}*sin\,\theta=m_b*a$$
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 not required further for this question, but will be important when we introduce friction on the surface of the incline plane.
$$a=\frac{F_{g}*sin\,\theta}{m_b}=\frac{m_b*g*sin\,\theta}{m_b}=g*sin\,\theta$$
$$a=9.81{m/s^2}*sin30=4.905{m/s^2}$$
The first step is to separate the two bodies in the diagram and assign the appropriate forces to each.
First, looking at Body 1, there are two forces and they are both acting in the \(y-axis\). As the question is stating that the block is stationary, the Net External Force is equal to zero as there is no acceleration.
$$\sum{F_{y}}=T-F_{g}=0,\;T=F_{g}$$
Moving to Body 2 (the pulley), pulleys redirect forces into another direction; therefore, the tension that the block pulls downward is equivalent to the applied force being pulled to the left.
$$F_a=T$$
Combining the two equations, we are left with
$$F_a=F_g=m_bg=25kg*9.81m/s^2=245N$$
The first step is to separate the three bodies in the diagram and assign the appropriate forces to each.
First, looking at Body 1, there are two forces and they are both acting in the \(y-axis\). As the question is stating that the block is stationary, the Net External Force is equal to zero as there is no acceleration.
$$\sum{F_{y}}=T_2-F_{g}=0,\;T_2=F_g$$
Moving to Body 2 (the suspended pulley), pulleys redirect forces into another direction around the wheel portion. In this body, the wheel itself has force applied in the downward direction and the tension on the rope is upward on both ends of the rope. Again, as the system is stationary, the Net External Force is equal to zero.
$$\sum{F_{y}}=T_1+T_1-T_2=0,\;T_1=\frac{1}{2}T_2$$
$$T_1=\frac{1}{2}F_g=\frac{1}{2}m_bg$$
Body 3 is the same as the previous question where the applied force is equivalent to the tension \(F_a=T_1\). Combining the three sets of equations we get:
\[F_a=T_1=\frac{1}{2}m_bg=\frac{1}{2}*25kg*9.81m/s^2\]
\[F_a=123N\]
The force required to keep the block stationary is 123N.
Note that this is half the force of the previous example. This is why pulleys are used to gain advantage in lifting systems.