Chapter 4, Problem 101QAP

To determine

The acceleration of the masses and tension in the strings.

Answer to Problem 101QAP

Tension in String ${\text{T}}_1=8.40\text{N}$

Tension in String ${\text{T}}_2=25.2\text{N}$

acceleration of the masses$a=3.50{\text{m/s}}^{\text{2}}$

Explanation of Solution

Givendata:

Mass of block A, $m_{1-\text{kg}}=1.00\text{kg}$

Mass of block B, $m_{2-\text{kg}}=2.00\text{kg}$

Mass of block B, $m_{4-\text{kg}}=4.00\text{kg}$

Formula Used:

Newton's second law:

  $F_{\text{net}}=m\times a$

Calculation:

For theblocks on the ramp, we can use a coordinate system wherethe axes are parallel and perpendicular to the face of theramp and up the ramp is considered positive.

For the $4.00\text{kg}$ block, down will be positive yin order to stay consistent with the ramp's coordinate system.

All three blocks have the sameacceleration because they are joined together.

Using Newton's second law for each of theblocks, we can solve for the acceleration in terms of the known masses.

Once we calculate theacceleration, we can calculate the magnitudes of the tensions.

  

Free-body diagram of $1.00\text{kg}$ block:

  

Free-body diagram of $2.00\text{kg}$block:

  

Free-body diagram of $4.00\text{kg}$ block:

  

Newton's second law:

  $\begin{array}{l}\sum F_{\text{ext,1-kg,parallel}}\text{=}{T}_1-w_{1-\text{kg}}=T_1-m_{1-\text{kg}}g\sin ({30.0}^{\circ })=m_{1-\text{kg}}a\\ \sum F_{\text{ext,2-kg,parallel}}{\text{=T}}_2-T_1-w_{2-\text{kg}}=T_2-T_1-m_{2-\text{kg}}g\sin ({30.0}^{\circ })=m_{2-\text{kg}}a\\ \sum F_{\text{ext,4-kg,y}}\text{=}-{\text{T}}_2+w_{4-\text{kg}}=-T_2+w_{4-\text{kg}}=-T_2+m_{4-\text{kg}}g=m_{4-\text{kg}}a\end{array}$

Solving for the acceleration:

  $\begin{array}{l}{\text{T}}_2=m_{4-\text{kg}}(g-a)=\left(m_{1-\text{kg}}+m_{2-\text{kg}}\right)a+\left(m_{1-\text{kg}}+m_{2-\text{kg}}\right)g\sin ({30.0}^{\circ })\\ a=\frac{m_{4-\text{kg}}g-\left(m_{1-\text{kg}}+m_{2-\text{kg}}\right)g\sin ({30.0}^{\circ })}{\left(m_{1-\text{kg}}+m_{2-\text{kg}}+m_{4-\text{kg}}\right)}\\ =\frac{\left(4.00\text{kg}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)-\left(\left(1.00\text{kg}\right)+\left(2.00\text{kg}\right)\right)\left(9.80{\text{m/s}}^{\text{2}}\right)\sin ({30.0}^{\circ })}{\left(1.00\text{kg}\right)+\left(2.00\text{kg}\right)+\left(4.00\text{kg}\right)}=3.50{\text{m/s}}^{\text{2}}\end{array}$

Solving for the tensions:

  $\begin{array}{l}{\text{T}}_1=m_{1-\text{kg}}g\sin ({30.0}^{\circ })+m_{1-\text{kg}}a=\left(1.00\text{kg}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)\sin ({30.0}^{\circ })+\left(1.00\text{kg}\right)\left(3.50{\text{m/s}}^{\text{2}}\right)=8.40\text{N}\\ {\text{T}}_2=m_{4-\text{kg}}(g-a)=\left(4.00\text{kg}\right)\left(\left(9.80{\text{m/s}}^{\text{2}}\right)-\left(3.50{\text{m/s}}^{\text{2}}\right)\right)=25.2\text{N}\end{array}$

Conclusion:

Thus, we havethe acceleration of the masses and tension in the strings as below: -

Tension in String ${\text{T}}_1=8.40\text{N}$

Tension in String ${\text{T}}_2=25.2\text{N}$

acceleration of the masses $a=3.50{\text{m/s}}^{\text{2}}$