Chapter 5, Problem 83QAP

To determine

The range of the values of the mass m₂ so that the two connected blocks resting on two inclined planes are in equilibrium.

Answer to Problem 83QAP

The mass m₂ can have values between 3.51 kg and 52.64 kg for the system of masses to be in equilibrium.

Explanation of Solution

Given info:

  $\begin{array}{l}{m}_1=6.00\text{ kg}\\ {\theta }_1=60°\\ {\theta }_2=35°\\ {\mu }_s=0.542\end{array}$

Formula used:

From the diagram given, if the value of m₂ falls below a certain value, it would slide up the plane and the block of mass m₁ would slide down. This gives the value of the minimum mass of m₂.

If the value of m₂ increases beyond a certain value, the block would slide down the plane and the block of mass m₁ would slide up.

By drawing the free body diagrams for each case, and applying the force equations for each case, the range of the values of m₂ can be determined.

Explanation and Calculation:

Case 1:

Consider the case when the block 1slides down the plane and the block 2 slides up the plane.

The free body diagram for m₁ is shown below:

The block rests on an incline which makes an angle ${\theta }_1$ to the horizontal. Assume a coordinate system which has the +x direction pointing down the plane and the +y direction perpendicular to the plane away from it.

  

The weight $w_1=m_{1}g$ of the block acts vertically downwards. The tension T acts along the −x direction, while the normal force n₁ acts along the +y direction. Since the block tends to slide down, the force of static friction f_s1 acts along the −x direction.

Resolve the weight w₁ into components along the +x and −y directions as shown. The magnitudes of the components are given by,

  $\begin{array}{l}{w}_{1x}=w_1\sin {\theta }_1=m_{1}g\sin {\theta }_1\\ w_{1y}=w_1\cos {\theta }_1=m_{1}g\cos {\theta }_1\end{array}$......(1)

Write the force equation along the +y direction.

  $\begin{array}{c}{\displaystyle \sum F_{1y}}=n_1-w_{1y}\\ =n_1-m_{1}g\cos {\theta }_1\end{array}$

Since the block is in equilibrium,

  ${\displaystyle \sum F_{1y}}=n_1-m_{1}g\cos {\theta }_1=0$

Therefore,

  $n_1=m_{1}g\cos {\theta }_1$

The force of friction is given by,

  $f_{s1}={\mu }_s{n}_1={\mu }_s{m}_{1}g\cos {\theta }_1$......(2)

The system is in equilibrium along the x direction too. Therefore,

  ${\displaystyle \sum F_{1x}}=w_{1x}-T-f_{s1}=0$

Using equations (1) and (2), in the above equation,

  $m_{1}g\sin {\theta }_1-T-{\mu }_s{m}_{1}g\cos {\theta }_1=0$......(3)

In a similar manner, construct a free body diagram for the block 2.

The block rests on an incline which makes an angle ${\theta }_2$ to the horizontal. Assume a coordinate system which has the +x direction pointing up the plane and the +y direction perpendicular to the plane away from it.

  

The weight $w_2=m_{2}g$ of the block acts vertically downwards. The tension T acts along the +x direction, while the normal force n₂ acts along the +y direction. Since the block tends to slide up, the force of static friction f_s2 acts along the −x direction.

Resolve the weight w₂ into components along the −x and −y directions as shown. The magnitudes of the components are given by,

  $\begin{array}{c}{w}_{2x}=w_2\sin {\theta }_2=m_{2}g\sin {\theta }_2\\ w_{2y}=w_2\cos {\theta }_2=m_{2}g\cos {\theta }_2\end{array}$......(4)

Write the force equation along the +y direction and apply the condition for equilibrium.

  $\begin{array}{c}{\displaystyle \sum F_{2y}}=n_2-w_{2y}=0\\ =n_2-m_{2}g\cos {\theta }_2=0\end{array}$

Therefore,

  $n_2=m_{2}g\cos {\theta }_2$......(5)

The force of friction is given by,

  $f_{s2}={\mu }_s{n}_2={\mu }_s{m}_{2}g\cos {\theta }_2$......(6)

The system is in equilibrium along the x direction too. Therefore,

  ${\displaystyle \sum F_{2x}}=T-w_{2x}-f_{s2}=0$

Using equations (4) and (6), in the above equation,

  $T-m_{2}g\sin {\theta }_2-{\mu }_s{m}_{2}g\cos {\theta }_2=0$......(7)

Add equations (3) and (7) and write an expression for m₂.

  $\begin{array}{c}{m}_{1}g\sin {\theta }_1-T-{\mu }_s{m}_{1}g\cos {\theta }_1=0\\ T-m_{2}g\sin {\theta }_2-{\mu }_s{m}_{2}g\cos {\theta }_2=0\end{array}$

  $\begin{array}{c}{m}_{2}g\left(\sin {\theta }_2+{\mu }_s\cos {\theta }_2\right)=m_{1}g\left(\sin {\theta }_1-{\mu }_s\cos {\theta }_1\right)\\ m_2=\frac{\left(\sin {\theta }_1-{\mu }_s\cos {\theta }_1\right)}{\left(\sin {\theta }_2+{\mu }_s\cos {\theta }_2\right)}{m}_1\end{array}$

Substitute the known values of the variables in the expression and calculate the value of m₂.

  $\begin{array}{c}{m}_2=\frac{\left(\sin {\theta }_1-{\mu }_s\cos {\theta }_1\right)}{\left(\sin {\theta }_2+{\mu }_s\cos {\theta }_2\right)}{m}_1\\ =\frac{\sin 60°-\left(0.542\right)\left(\cos 60°\right)}{\sin 35°+\left(0.542\right)\left(\cos 35°\right)}\left(6.00\text{ kg}\right)\\ =3.51\text{ kg}\end{array}$

The minimum value of m₂ for which the system is in equilibrium is 3.51 kg.

Case 2:

Consider the case when the block 1slides up the plane and the block 2 slides down the plane.

The free body diagram for m₁ is shown below:

Assume the +x direction up the incline and the +y direction perpendicular to the plane. The weight $w_1=m_{1}g$ of the block acts vertically downwards. The tension T acts along the +x direction, while the normal force n₁ acts along the +y direction. Since the block tends to slide up, the force of static friction f_s1 acts along the −x direction.

  

The equations (1) and (2) are applied to this free body diagram too.

Write the condition for equilibrium along the x direction.

  ${\displaystyle \sum F_{1x}}=T-w_{1x}-f_{s1}=0$

Use equations (1) and (2) in the expression.

  $T-m_{1}g\sin {\theta }_1-{\mu }_s{m}_{1}g\cos {\theta }_1=0$......(8)

In a similar manner, construct a free body diagram for the block 2.

. Assume a coordinate system which has the +x direction pointing down the plane and the +y direction perpendicular to the plane away from it. The weight $w_2=m_{2}g$ of the block acts vertically downwards. The tension T acts along the −x direction, while the normal force n₂ acts along the +y direction. Since the block tends to slide down, the force of static friction f_s2 acts along the −x direction.

  

The equations (4), (5) and (6) are valid for this case too.

Write the equation for equilibrium along the x direction.

  ${\displaystyle \sum F_{2x}}=w_{2x}-T-f_{s2}=0$

Using equations (4) and (6) in the expression,

  $m_{2}g\sin {\theta }_2-T-{\mu }_s{m}_{2}g\cos {\theta }_2=0$......(9)

Add equations (8) and (9) and write an expression for m₂.

  $\begin{array}{c}T-m_{1}g\sin {\theta }_1-{\mu }_s{m}_{1}g\cos {\theta }_1=0\\ m_{2}g\sin {\theta }_2-T-{\mu }_s{m}_{2}g\cos {\theta }_2=0\end{array}$

  $m_{2}g\left(\sin {\theta }_2-{\mu }_s\cos {\theta }_2\right)=m_{1}g\left(\sin {\theta }_1+{\mu }_s\cos {\theta }_1\right)$

  $m_2=\frac{\left(\sin {\theta }_1+{\mu }_s\cos {\theta }_1\right)}{\left(\sin {\theta }_2-{\mu }_s\cos {\theta }_2\right)}{m}_1$

Substitute the known values of the variables in the expression.

  $\begin{array}{c}{m}_2=\frac{\left(\sin {\theta }_1+{\mu }_s\cos {\theta }_1\right)}{\left(\sin {\theta }_2-{\mu }_s\cos {\theta }_2\right)}{m}_1\\ =\frac{\sin 60°+\left(0.542\right)\left(\cos 60°\right)}{\sin 35°-\left(0.542\right)\left(\cos 35°\right)}\left(6.00\text{ kg}\right)\\ =52.64\text{ kg}\end{array}$

The maximum value of m₂ for the system to be in equilibrium is 52.64 kg.

Conclusion:

Thus, the mass m₂ can have values between 3.51 kg and 52.64 kg for the system of masses to be in equilibrium.