Chapter 5, Problem 43P

Consider the three connected objects shown in Figure P5.43. Assume first that the inclined plane is frictionless and that the system is in equilibrium. In terms of m, g, and θ, find (a) the mass M and (b) the tensions T₁ and T₂. Now assume that the value of M is double the value found in part (a). Find (c) the acceleration of each object and (d) the tensions T₁ and T₂. Next, assume that the coefficient of static friction between m and 2m and the inclined plane is μ_s and that the system is in equilibrium. Find (e) the maximum value of M and (f) the minimum value of M. (g) Compare the values of T₂ when M has its minimum and maximum values.

Figure P5.43

To determine

The mass $M$ in terms of $m$, $g$, and $\theta$.

Answer to Problem 43P

The mass $M$ in terms of $m$, $g$, and $\theta$ is $\underset{\_}{3m\sin \theta }$.

Explanation of Solution

The free body diagram of the mass $2m$ is shown in Figure 1.

The free body diagram of the mass $m$ is shown in Figure 2.

The free body diagram of the mass $M$ is shown in Figure 3.

Apply Newton’s second law in the Figure1.

    $T_1=f_1+2m\left(g\sin \theta +a\right)$        (I)

Here, $T_1$ is the tension acting on the mass $2m$, $m$ is the mass, $g$ is the acceleration due to gravity, $f_1$  frictional force on the mass $2m$, $a$ is the acceleration, and $\theta$ is the angle of inclination.

Apply Newton’s second law in the Figure 2.

    $T_2-T_1=f_2+m\left(g\sin \theta +a\right)$        (II)

Here, $T_2$ is the tension acting on the mass $m$, and $f_2$ is the frictional force on mass $m$.

Apply Newton’s second law in the Figure 3.

    $T_2=M\left(g-a\right)$        (III)

Assume that, the system is in equilibrium, $a=0$, and the incline is friction less $f_1=f_2=0$.

Use equation (I) and (III), in (II).

    $\begin{array}{c}Mg-2mg\sin \theta =mg\sin \theta \\ M=3m\sin \theta \end{array}$        (IV)

Conclusion:

Therefore, the mass $M$ in terms of $m$, $g$, and $\theta$ is $\underset{\_}{3m\sin \theta }$.

To determine

The tensions $T_1$ and $T_{\text{2}}$.

Answer to Problem 43P

The tension $T_1$ is $\underset{\_}{2mg\sin \theta }$, and $T_{\text{2}}$ is $\underset{\_}{3mg\sin \theta }$.

Explanation of Solution

From the part (a), $T_1$ is $2mg\sin \theta$ when the system is in equilibrium.

Use equation (IV) in (III), and substitute $0$ for $a$ and $f_1$.

    $T_2=3mg\sin \theta$

Conclusion:

Therefore, the tensions $T_1$ is $\underset{\_}{2mg\sin \theta }$, and $T_{\text{2}}$ is $\underset{\_}{3mg\sin \theta }$.

To determine

The acceleration of each object when the value of $M$ is doubled.

Answer to Problem 43P

The acceleration of each object when the value of $M$ is doubled is $\underset{\_}{a=\frac{g\sin \theta }{1+2\sin \theta }}$.

Explanation of Solution

Assume that, $M=6m\sin \theta$, and $f_1=f_2=0$.

Substitute, $6m\sin \theta$ for $M$ in the equation (III).

    $T_2=6m\sin \theta \left(g-a\right)$        (V)

Use equation (I) and (V) in (VII), substitute, $0$ for $f_1$, and $f_2$, and solve for $a$.

    $\begin{array}{c}6m\sin \theta \left(g-a\right)-2m\left(g\sin \theta +a\right)=m\left(g\sin \theta +a\right)\\ 6mg\sin \theta -6m\sin \theta a-2mg\sin \theta -2ma=mg\sin \theta +ma\\ 3g\sin \theta -3a-6\sin \theta a=0\\ a=\frac{3g\sin \theta }{3+6\sin \theta }\\ =\frac{g\sin \theta }{1+2\sin \theta }\end{array}$        (VI)

Conclusion:

Therefore, the acceleration of each object when the value of $M$ is doubled is $\underset{\_}{a=\frac{g\sin \theta }{1+2\sin \theta }}$.

To determine

The tensions $T_1$, and $T_2$ when the value of $M$ is doubled.

Answer to Problem 43P

The tensions $T_1$ is $\underset{\_}{T_1=4mg\sin \theta \left(\frac{1+\sin \theta }{1+2\sin \theta }\right)}$ and $T_2$ is $\underset{\_}{T_2=6mg\sin \theta \left(\frac{1+\sin \theta }{1+2\sin \theta }\right)}$ when the value of $M$ is doubled.

Explanation of Solution

Use equation (VI) in (I), substitute, $0$ for $f_1$.

    $\begin{array}{c}{T}_1=0+2m\left(g\sin \theta +\frac{g\sin \theta }{1+2\sin \theta }\right)\\ =2mg\sin \theta \left(1+\frac{1}{1+2\sin \theta }\right)\\ =2mg\sin \theta \left(\frac{1+2\sin \theta +1}{1+2\sin \theta }\right)\\ =4mg\sin \theta \left(\frac{1+\sin \theta }{1+2\sin \theta }\right)\end{array}$        (VII)

Use equation (VI) in (III), and use $6m\sin \theta$ for $M$.

    $\begin{array}{c}{T}_2=6m\sin \theta \left(g-\frac{g\sin \theta }{1+2\sin \theta }\right)\\ =6mg\sin \theta \left(1-\frac{\sin \theta }{1+2\sin \theta }\right)\\ =6mg\sin \theta \left(\frac{1+\sin \theta }{1+2\sin \theta }\right)\end{array}$        (VIII)

Conclusion:

Therefore, the tensions $T_1$ is $\underset{\_}{T_1=4mg\sin \theta \left(\frac{1+\sin \theta }{1+2\sin \theta }\right)}$ and $T_2$ is $\underset{\_}{T_2=6mg\sin \theta \left(\frac{1+\sin \theta }{1+2\sin \theta }\right)}$ when the value of $M$ is doubled.

To determine

The maximum value of $M$ at the equilibrium.

Answer to Problem 43P

The maximum value of $M$ at the equilibrium is $\underset{\_}{M_{\text{max}}=3m\left(\sin \theta +{\mu }_{\text{s}}\cos \theta \right)}$.

Explanation of Solution

At the equilibrium the acceleration is zero, and the frictional forces acting down the incline.

Use $2{\mu }_{\text{s}}mg\cos \theta$ for $f_1$, and $0$ for $a$ in the equation (I).

    $\begin{array}{c}{T}_1=2{\mu }_{\text{s}}mg\cos \theta +2mg\sin \theta \\ =2mg\left(\sin \theta +{\mu }_{\text{s}}\cos \theta \right)\end{array}$        (IX)

Use ${\mu }_{\text{s}}mg\cos \theta$ for $f_2$, and $0$ for $a$ in the equation (II).

    $\begin{array}{c}{T}_2-T_1={\mu }_{\text{s}}mg\cos \theta +mg\sin \theta \\ =mg\left(\sin \theta +{\mu }_{\text{s}}\cos \theta \right)\end{array}$        (X)

Rewrite the equation (III) for $M_{\text{max}}$.

    $T_{2,\text{max}}=M_{\text{max}}g$        (XI)

Use equation (IX) and (X) in (XI).

    $\begin{array}{c}{M}_{\text{max}}g-2mg\left(\sin \theta +{\mu }_{\text{s}}\cos \theta \right)=mg\left(\sin \theta +{\mu }_{\text{s}}\cos \theta \right)\\ M_{\text{max}}g=2mg\sin \theta +2mg{\mu }_{\text{s}}\cos \theta +mg\sin \theta +mg{\mu }_{\text{s}}\cos \theta \\ M_{\text{max}}=3m\sin \theta +3m{\mu }_{\text{s}}\cos \theta \\ =3m\left(\sin \theta +{\mu }_{\text{s}}\cos \theta \right)\end{array}$

Conclusion:

Therefore, the maximum value of $M$ at the equilibrium is $\underset{\_}{M_{\text{max}}=3m\left(\sin \theta +{\mu }_{\text{s}}\cos \theta \right)}$.

To determine

The minimum value of $M$ at the equilibrium.

Answer to Problem 43P

The minimum value of $M$ at the equilibrium is $\underset{\_}{M_{\text{min}}=3m\left(\sin \theta -{\mu }_{\text{s}}\cos \theta \right)}$.

Explanation of Solution

Rewrite the equation (III) for $M_{\text{min}}$.

    $T_{2,\text{min}}=M_{\text{min}}g$        (XII)

At the equilibrium the acceleration is zero, and impending motion up the incline, so the frictional forces acting up the incline.

Use $2{\mu }_{\text{s}}mg\cos \theta$ for $f_1$, and $0$ for $a$ in the equation (I).

    $\begin{array}{c}{T}_1=2{\mu }_{\text{s}}mg\cos \theta -2mg\sin \theta \\ =2mg\left(\sin \theta -{\mu }_{\text{s}}\cos \theta \right)\end{array}$        (XIII)

Since the frictional forces acting up the incline.

Use ${\mu }_{\text{s}}mg\cos \theta$ for $f_2$, and $0$ for $a$ in the equation (II).

    $\begin{array}{c}{T}_2-T_1={\mu }_{\text{s}}mg\cos \theta -mg\sin \theta \\ =mg\left(\sin \theta -{\mu }_{\text{s}}\cos \theta \right)\end{array}$        (XIV)

Since the frictional forces acting up the incline.

Use equation (XII) and (XIII) in (XIV).

    $\begin{array}{c}{M}_{\text{min}}g-2mg\left(\sin \theta -{\mu }_{\text{s}}\cos \theta \right)=mg\left(\sin \theta -{\mu }_{\text{s}}\cos \theta \right)\\ M_{\text{min}}g=2mg\sin \theta -2mg{\mu }_{\text{s}}\cos \theta +mg\sin \theta -mg{\mu }_{\text{s}}\cos \theta \\ M_{\text{min}}=3m\sin \theta -3m{\mu }_{\text{s}}\cos \theta \\ =3m\left(\sin \theta -{\mu }_{\text{s}}\cos \theta \right)\end{array}$        (XV)

Conclusion:

Therefore, the minimum value of $M$ at the equilibrium is $\underset{\_}{M_{\text{min}}=3m\left(\sin \theta -{\mu }_{\text{s}}\cos \theta \right)}$.

To determine

The value of $T_2$ when the mass $M$ has its maximum and minimum.

Answer to Problem 43P

The value of $T_2$ when the $M$ has its maximum and minimum is $\underset{\_}{6{\mu }_{\text{s}}mg\cos \theta }$.

Explanation of Solution

Subtract equation (XII) from (XI).

    $T_{\text{2,max}}-T_{\text{2,min}}=M_{\text{max}}g-M_{\text{min}}g$        (XVI)

Use $3m\left(\sin \theta +{\mu }_{\text{s}}\cos \theta \right)$ for $M_{\text{max}}$, and $3m\left(\sin \theta -{\mu }_{\text{s}}\cos \theta \right)$ for $M_{\text{min}}$ in the equation (XVI).

    $\begin{array}{c}{T}_{\text{2,max}}-T_{\text{2,min}}=3mg\left(\sin \theta +{\mu }_{\text{s}}\cos \theta \right)-3mg\left(\sin \theta -{\mu }_{\text{s}}\cos \theta \right)\\ =3mg{\mu }_{\text{s}}\cos \theta +3mg{\mu }_{\text{s}}\cos \theta \\ =6mg{\mu }_{\text{s}}\cos \theta \end{array}$

Conclusion:

Therefore, the value of $T_2$ when the $M$ has its maximum and minimum is $\underset{\_}{6{\mu }_{\text{s}}mg\cos \theta }$.