Solved Problem on Dynamics

A small block of mass m₁ is placed over another larger block of mass m₂, and this, on a horizontal plane. Block 1 is drawn with a force that makes an angle θ with the vertical, and bock 2 is drawn horizontally, the static friction coefficient between the blocks, and between the block and the plane is equal to μ. Determine the minimum values of the forces with which the blocks must be pulled so that the movement begins.

Problem data:

Mass of body 1: m₁;
Mass of body 2: m₂;
Angle of force 1: θ;
Coefficient of friction between blocks: μ;
Coefficient of friction between the block and the plane: μ.

Problem diagram:

Drawing free-bodies diagrams we have the forces in each block.

Block 1:

F₁: force with which block 1 is drawn;
F_at1: friction force between block 1 and block 2 due to force 1;
P₁: gravitational force on the block 1;
N₁: normal reaction force due to block 2.

Figure 1

Block 2:

F₂: force with which Block 2 is drawn;
F_at2: friction force between block 2 and plane;
F_at1: friction force between block 2 and block 1;
F_g2: gravitational force on the block 2;
N₁: normal reaction due to block 1,
N₂: normal reaction force with the plane.

Figure 2

Solution

As the blocks are initially at rest we have the condition that the sum of the forces that act on them is equal to zero

\[ \begin{gather} \bbox[#99CCFF,10px] {\sum {\mathbf{F}}=0} \tag{I} \end{gather} \]

applying this condition to block 1 (Figure 1-B)

\[ {\mathbf{F}}_{1}+{\mathbf{F}}_{at 1}+{\mathbf{N}}_{1}+{\mathbf{F}}_{g1}=0 \]

where

\( {\mathbf{F}}_{1}=F_{1}\sin \theta\;\mathbf{i}+F_{1}\cos \theta\;\mathbf{j} \qquad\text{(II)} \)
\( {\mathbf{F}}_{at 1}=-\mu N_{1}\;\mathbf{i} \)
\( {\mathbf{N}}_{1}=N_{1}\;\mathbf{j} \)
\( {\mathbf{f}}_{1}=-m_{g1}g\;\mathbf{j} \)

assim

\[ F_{1}\sin \theta \;\mathbf{i}+F_{1}\cos\theta \;\mathbf{j}-\mu N_{1}\;\mathbf{i}+N_{1}\;\mathbf{j}-m_{1}g\;\mathbf{j}=0 \tag{III} \]

Separating the components from the equation (III)

\[ \begin{gather} \left\{ \begin{array}{l} F_{1}\sin \theta -\mu N_{1}=0\\ F_{1}\cos \theta+N_{1}-m_{1}g=0 \end{array} \right. \tag{IV} \end{gather} \]

solving the second equation for N₁ in the second equation of system (IV) and substituting it into the first equation

\[ \begin{gather} N_{1}=m_{1}g-F_{1}\cos \theta\\[10pt] F_{1}\sin \theta =\mu \left(m_{1}g-F_{1}\cos \theta\right)\\ F_{1}\sin \theta =\mu m_{1}g-\mu F_{1}\cos \theta\\ F_{1}\sin \theta +\mu F_{1}\cos \theta =\mu m_{1}g\\ F_{1}\left(\sin \theta +\mu \cos \theta\right)=\mu m_{1}g\\ F_{1}=\frac{\mu m_{1}g}{\sin \theta+\mu \cos \theta } \end{gather} \]

Substituting this value in expression (II)) for vector F₁, so the movement begins the applied force should be higher than

\[ \bbox[#FFCCCC,10px] {{\mathbf{F}}_{1}=\frac{\mu m_{1}g}{\sin \theta +\mu \cos \theta}\left(\sin \theta \;\mathbf{i}+\cos \theta\;\mathbf{j}\right)} \]

Substituting F₁ in the second equation of system (IV), we find the normal reaction N₁

\[ \begin{gather} \frac{\mu m_{1}g}{\sin \theta +\mu \cos\theta }\cos \theta +N_{1}-m_{1}g=0\\[5pt] N_{1}=m_{1}g-\frac{\mu m_{1}g\cos\theta }{\sin \theta +\mu \cos \theta}\\[5pt] N_{1}=m_{1}g-\frac{\mu m_{1}g\cos \theta }{\sin \theta\left(1+\mu \dfrac{\cos \theta}{\sin \theta}\right)}\\[5pt] N_{1}=m_{1}g-\frac{\mu m_{1}g\cot \theta}{1+\mu \cot \theta} \tag{V} \end{gather} \]

Applying the condition (I) to block 2 (Figure 2-B)

\[ {\mathbf{F}}_{2}+{\mathbf{F}}_{at 2}+{\mathbf{N}}_{2}+{\mathbf{F}}_{g2}+{\mathbf{N}}_{1}+{\mathbf{F}}_{at 1}=0 \]

where

\( {\mathbf{F}}_{2}=-F_{2}\;\mathbf{i} \qquad\text{(VI)} \)
\( {\mathbf{F}}_{at 2}=\mu N_{2}\;\mathbf{i} \)
\( {\mathbf{N}}_{1}=-N_{1}\;\mathbf{j} \)
\( {\mathbf{N}}_{2}=N_{2}\;\mathbf{j} \)
\( {\mathbf{F}}_{g2}=-m_{2}g\;\mathbf{j} \)
\( {\mathbf{F}}_{at 1}=\mu N_{1}\;\mathbf{i} \)

so

\[ -F_{2}\;\mathbf{i}+\mu N_{2}\;\mathbf{i}-N_{1}\;\mathbf{j}+N_{2}\;\mathbf{j}-m_{2}g\;\mathbf{j}+\mu N_{1}\;\mathbf{i}=0 \tag{VII} \]

Separating the components from the equation (VII)

\[ \begin{gather} \left\{ \begin{array}{l} -F_{2}+\mu N_{2}+\mu N_{1}=0\\ -N_{1}+N_{2}-m_{2}g=0 \end{array} \right. \tag{VIII} \end{gather} \]

solving the second equation for N₂ in the system (VIII) and substituting it into the first equation

\[ \begin{gather} N_{2}=N_{1}+m_{2}g\\[10pt] -F_{2}+\mu\left(N_{1}+m_{2}g\right)+\mu N_{1}=0\\ F_{2}=\mu N_{1}+\mu m_{2}g+\mu N_{1}\\ F_{2}=2\mu N_{1}+\mu m_{2}g \tag{IX} \end{gather} \]

substituting the expression (V) into expression (IX)

\[ \begin{gather} F_{2}=\mu \left[m_{2}g+2\left(m_{1}g-\frac{\mu m_{1}g\cot \theta }{1+\mu \cot \theta}\right)\right]\\[5pt] F_{2}=\mu \left[m_{2}g+2m_{1}g-2\frac{\mu m_{1}g\cot \theta }{1+\mu \cot \theta}\right] \end{gather} \]

Substituting this value in the expression (VI) for the vector F₂, so that the movement begins the applied force should be higher than

\[ \bbox[#FFCCCC,10px] {{\mathbf{F}}_{2}=-\mu\left[\left(m_{2}+2m_{1}\right)g-2\frac{\mu m_{1}g\cot \theta }{1+\mu \cot \theta}\right]\;\mathbf{i}} \]

Note: See that in particular, if \( \theta=\frac{\pi}{2} \) solutions for F₁ and F₂ forces are reduced to

\[ \begin{gather} \mathbf{F}_{1}=\frac{\mu m_{1}g}{\cancelto{1}{\sin \frac{\pi}{2}} +\mu \cancelto{0}{\cos \frac{\pi}{2}}}\left(\cancelto{1}{\sin \frac{\pi}{2}} \;\mathbf{i}+\cancelto{0}{\cos \frac{\pi}{2}}\;\mathbf{j}\right)\\[5pt] \mathbf{F}_{1}=\mu m_{1}g \;\mathbf{i} \end{gather} \]

that is the force of friction, then the force F₁ should be greater than this for the movement to start.

\[ \begin{gather} \mathbf{F}_{2}=-\mu\left[\left(m_{2}+2m_{1}\right)g-2\frac{\mu m_{1}g \frac{\cancelto{0}{\cos \theta}}{\cancelto{1}{\sin \theta}} }{1+\mu \frac{\cancelto{0}{\cos \theta}}{\cancelto{1}{\sin \theta}}}\right]\;\mathbf{i}\\[5pt] \mathbf{F}_{2}=-\mu\left(m_{2}+2m_{1}\right)g\\[5pt] \mathbf{F}_{2}=-\mu\left(m_{2}+m_{1}+m_{1}\right)g\\[5pt] \mathbf{F}_{2}=-\mu\left(m_{2}+m_{1}\right)g+\mu m_{1}g \end{gather} \]

which is the force of friction of the whole system (m₁+m₂) on the ground. The force of friction between the blocks (depends on m₁), then the force F₂ should be greater than this for the movement to begin.

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .