Chapter 4, Problem 74QAP

To determine

(a)

The value of $M_1$ for which the two blocks are in equilibrium (no acceleration)

Answer to Problem 74QAP

The value of $M_1=10.0\text{kg}$ for which the two blocks are in equilibrium (no acceleration)

Explanation of Solution

Given data:

  $M_2=20.0\text{kg}$

  $\theta ={30.0}^{\circ }$

Formula Used:

Newton's second law:

  $F_{\text{net}}=m\times a$

  $M(\text{Mass})=\frac{T(\text{Force)}}{g(\text{acceleration)}}$

Calculation:

  

We'll use two different but related coordinate systems for the two blocks. For block $M_1$, positive y will point upward. For block $M_2$, the axes will be parallel and perpendicular to the inclined plane, where down the ramp and out of the ramp are positive.

We can draw free-body diagrams and apply Newton's second law in component form for both blocks.

The tension acting on each block will be identical in magnitude.

Since block $M_2$ does not leave the plane, the acceleration in the perpendicular direction is always zero.

Since, both blocks are at rest, which means the acceleration of each block is zero and we can calculate $M_1$ in this case.

.

Free-body diagram of $M_2$ :

  

Newton's second law for $M_2$ :

  $\begin{array}{l}\sum F_{\text{ext}\text{,parallel}}=-T+w_{M_2,\text{parallel}}=M_2{a}_{\text{parallel}}=0\\ T=w_{M_2,\text{parallel}}=M_{2}g\sin ({30.0}^{\circ })=(20.0\text{kg)(9}\text{.80}{\text{m/s}}^2)\sin ({30.0}^{\circ })=98.0\text{N}\end{array}$

Free-body diagram of $M_1$ :

  

  $\begin{array}{l}\sum F_{\text{ext}\text{,y}}=T-w_{M_1}=M_1{a}_y=0\\ =>w_{M_1}=M_{1}g=T\\ =>M_1=\frac{T}{g}=\frac{98.0\text{N}}{\left(9.80{\text{m/s}}^{\text{2}}\right)}=10.0\text{kg}\end{array}$

Conclusion:

For two blocks to be in equilibrium the value of $M_1=10.0\text{kg}$.

To determine

(b)

The magnitude of acceleration of two blocks if the system can move.

Answer to Problem 74QAP

The magnitude of acceleration of two blocks $1.96{\text{m/s}}^2$.

Explanation of Solution

Given data:

  $M_1=5.0\text{kg}$

  $M_2=20.0\text{kg}$

  $\theta ={30.0}^{\circ }$

Formula Used:

Newton's second law:

  $F_{\text{net}}=m\times a$

  $M(\text{Mass})=\frac{T(\text{Force)}}{g(\text{acceleration)}}$

Calculation:

  

We'll use two different but related coordinate systems for the two blocks. For block $M_1$, positive y will point upward. For block $M_2$, the axes will be parallel and perpendicular to the inclined plane, where down the ramp and out of the ramp are positive.

We can draw free-body diagrams and apply Newton's second law in component form for both blocks.

The tension acting on each block will be identical in magnitude.

Since block $M_2$ does not leave the plane, the acceleration in the perpendicular direction is always zero.

Since, $M_1=5.00\text{kg}$, so we can plug this into Newton's second law and calculate the acceleration of the blocks.

The magnitude of the acceleration will be the same for each block because they are joined together with a string.

Free-body diagram of $M_1$ :

  

  $\begin{array}{l}\sum F_{\text{ext}\text{,y}}=T-w_{M_1}=T-M_{1}g=M_1{a}_y=M_{1}a\\ =>T=M_1(g+a)\end{array}$

Free-body diagram of $M_2$ :

  

Newton's second law for $M_2$ :

  $\begin{array}{l}\sum F_{\text{ext}\text{,parallel}}=-T+w_{M_2,\text{parallel}}=-(M_1(g+a))+M_{2}g\sin ({30.0}^{\circ })=M_2{a}_{\text{parallel}}=M_{2}a\\ =>a=\frac{(M_2\sin ({30.0}^{\circ })-M_1)g}{M_1+M_2}=\frac{((20.0\text{kg})\sin ({30.0}^{\circ })-(5.0\text{kg)})\text{(9}\text{.80}{\text{m/s}}^2)}{(20.0\text{kg})+(5.0\text{kg)}}=1.96{\text{m/s}}^2\end{array}$

Conclusion:

If the system can move the magnitude of acceleration of two blocks is $1.96{\text{m/s}}^2$

To determine

(c)

The direction of $M_2$.

Answer to Problem 74QAP

The direction of $M_2$is down the ramp.

Explanation of Solution

Given data:

  $M_1=5.0\text{kg}$

  $M_2=20.0\text{kg}$

  $\theta ={30.0}^{\circ }$

Formula Used:

Newton's second law:

  $F_{\text{net}}=m\times a$

Calculation:

  

We'll use two different but related coordinate systems for the two blocks. For block $M_1$, positive y will point upward. For block $M_2$, the axes will be parallel and perpendicular to the inclined plane, where down the ramp and out of the ramp are positive.

We can draw free-body diagrams and apply Newton's second law in component form for both blocks.

The tension acting on each block will be identical in magnitude.

Since block $M_2$ does not leave the plane, the acceleration in the perpendicular direction is always zero.

Since, $M_1=5.00\text{kg}$, so we can plug this into Newton's second law and calculate the acceleration of the blocks. The magnitude of the acceleration will be the same for each block because they are tethered together with a string.

Free-body diagram of $M_1$ :

  

  $\begin{array}{l}\sum F_{\text{ext}\text{,y}}=T-w_{M_1}=T-M_{1}g=M_1{a}_y=M_{1}a\\ =>T=M_1(g+a)\end{array}$

Free-body diagram of $M_2$ :

  

Newton's second law for $M_2$ :

  $\begin{array}{l}\sum F_{\text{ext}\text{,parallel}}=-T+w_{M_2,\text{parallel}}=-(M_1(g+a))+M_{2}g\sin ({30.0}^{\circ })=M_2{a}_{\text{parallel}}=M_{2}a\\ =>a=\frac{(M_2\sin ({30.0}^{\circ })-M_1)g}{M_1+M_2}=\frac{((20.0\text{kg})\sin ({30.0}^{\circ })-(5.0\text{kg)})\text{(9}\text{.80}{\text{m/s}}^2)}{(20.0\text{kg})+(5.0\text{kg)}}=1.96{\text{m/s}}^2\end{array}$

Since $a=1.96{\text{m/s}}^2$ is positive, $M_2$ moves down the ramp.

Conclusion:

As we have value of $a$ as positive, so, $M_2$ moves down the ramp.

To determine

(d)

The distance to which block $M_2$ move in $2.00\text{s}$.

Answer to Problem 74QAP

The distance to which block $M_2$move in $2.00\text{s}$is $3.92\text{m}$

Explanation of Solution

Given data:

  $M_1=5.0\text{kg}$

  $M_2=20.0\text{kg}$

  $\theta ={30.0}^{\circ }$

Formula Used:

Newton's second law:

  $F_{\text{net}}=m\times a$

And $x=ut+\frac{1}{2}a{t}^2$

Calculation:

  

We'll use two different but related coordinate systems for the two blocks. For block $M_1$, positive y will point upward. For block $M_2$, the axes will be parallel and perpendicular to the inclined plane, where down the ramp and out of the ramp are positive.

We can draw free-body diagrams and apply Newton's second law in component form for both blocks.

The tension acting on each block will be identical in magnitude.

Since block $M_2$ does not leave the plane, the acceleration in the perpendicular direction is always zero.

Since, $M_1=5.00\text{kg}$, so we can plug this into Newton's second law and calculate the acceleration of the blocks. The magnitude of the acceleration will be the same for each block because they are tethered together with a string.

If the acceleration is constant, we can use the constant acceleration equations to calculate how far $M_2$ travels starting from rest.

Free-body diagram of $M_1$ :

  

  $\begin{array}{l}\sum F_{\text{ext}\text{,y}}=T-w_{M_1}=T-M_{1}g=M_1{a}_y=M_{1}a\\ =>T=M_1(g+a)\end{array}$

Free-body diagram of $M_2$ :

  

Newton's second law for $M_2$ :

  $\begin{array}{l}\sum F_{\text{ext}\text{,parallel}}=-T+w_{M_2,\text{parallel}}=-(M_1(g+a))+M_{2}g\sin ({30.0}^{\circ })=M_2{a}_{\text{parallel}}=M_{2}a\\ =>a=\frac{(M_2\sin ({30.0}^{\circ })-M_1)g}{M_1+M_2}=\frac{((20.0\text{kg})\sin ({30.0}^{\circ })-(5.0\text{kg)})\text{(9}\text{.80}{\text{m/s}}^2)}{(20.0\text{kg})+(5.0\text{kg)}}=1.96{\text{m/s}}^2\end{array}$

By using the constant acceleration equations to calculate how far $M_2$ travels starting from rest.

  $x=x_0+v_{0,\text{parallel}}t+\frac{1}{2}{a}_{\text{parallel}}{t}^2=0+0+\frac{1}{2}(1.96{\text{m/s}}^2){(2.00\text{s)}}^2=3.92\text{m}$

Conclusion:

Thus in $2.00\text{s}$ block $M_2$ moves $3.92\text{m}$ of distance.