Chapter 4, Problem 61QAP

To determine

(a)

The force that the table exerts on box A if box B weighs $35\text{}\text{N}$

Answer to Problem 61QAP

The force exerted by table on box A is $=45\text{N}$ and vertically upwards.

Explanation of Solution

Given:

  

Box A weighs $=80\text{N}$

Box B weighs $=35\text{N}$

Pulley and rope are massless.

Formula Used:

  $\begin{array}{l}{\displaystyle \sum F_{\text{ext,Ay}}=T+n-w_{\text{A}}}=m_{\text{A}}{a}_{\text{y}}=0\\ =>n=w_{\text{A}}-T\end{array}$

Calculation:

The forces acting on Box A are the tension in the rope, the force due to gravity, and the force the table exerts on box A (also known as the normal force).

The tension in the rope is equal to the weight of box B when box B is at rest.

If the weight of box B is larger than the weight of box A, both boxes start to move-box A moves up, while box B moves down.

Free-body diagram of box A:

  

If box A remains at rest, then

  $\begin{array}{l}{\displaystyle \sum F_{\text{ext,Ay}}=T+n-w_{\text{A}}}=m_{\text{A}}{a}_{\text{y}}=0\\ =>n=w_{\text{A}}-T\end{array}$

If box B weighs $35\text{}\text{N}$, then the tension $T=35\text{}\text{N}$, since Box B remains at rest.

  $=>n=w_{\text{A}}-T=(80\text{N})-(35\text{}\text{N})=45\text{N}$ and points up

Conclusion:

The force that table exerts on box A if box B weighs $35\text{N}$ is $=45\text{N}$ and points up.

To determine

(b)

The force that the table exerts on box A if box B weighs $70\text{}\text{N}$

Answer to Problem 61QAP

The force exerted by table on box A is $\text{=}\text{10}\text{N}$ and vertically upwards.

Explanation of Solution

Given info:

  

Box A weighs $=80\text{N}$

Box B weighs $=70\text{}\text{N}$

Pulley and rope are massless.

Formula Used:

  $\begin{array}{l}{\displaystyle \sum F_{\text{ext,Ay}}=T+n-w_{\text{A}}}=m_{\text{A}}{a}_{\text{y}}=0\\ =>n=w_{\text{A}}-T\end{array}$

Calculation:

The forces acting on Box A are the tension in the rope, the force due to gravity, and the force the table exerts on box A (also known as the normal force).

The tension in the rope is equal to the weight of box B when box B is at rest.

If the weight of box B is larger than the weight of box A, both boxes start to move-box A moves up, while box B moves down.

Free-body diagram of box A:

  

If box A remains at rest, then

  $\begin{array}{l}{\displaystyle \sum F_{\text{ext,Ay}}=T+n-w_{\text{A}}}=m_{\text{A}}{a}_{\text{y}}=0\\ =>n=w_{\text{A}}-T\end{array}$

If box B weighs $70\text{}\text{N}$, then the tension $T=70\text{}\text{N}$, since Box B remains at rest.

  $=>n=w_{\text{A}}-T=(80\text{N})-(70\text{}\text{N})=10\text{N}$ and points up

Conclusion:

The force that table exerts on box A if box B weighs $70\text{}\text{N}$ is $\text{=}\text{10}\text{N}$ and points up.

To determine

(c)

The force that the table exerts on box A if box B weighs $90\text{N}$

Answer to Problem 61QAP

The force that table exerts on box A if box B weighs $90\text{N}$ is $=0\text{}\text{N}$.

Explanation of Solution

Given:

  

Box A weighs $=80\text{N}$

Box B weighs $=90\text{N}$

Pulley and rope are massless.

Formula Used:

  $\begin{array}{l}{\displaystyle \sum F_{\text{ext,Ay}}=T+n-w_{\text{A}}}=m_{\text{A}}{a}_{\text{y}}=0\\ =>n=w_{\text{A}}-T\end{array}$

Calculation:

The forces acting on Box A are the tension in the rope, the force due to gravity, and the force the table exerts on box A (also known as the normal force).

The tension in the rope is equal to the weight of box B when box B is at rest.

If the weight of box B is larger than the weight of box A, both boxes start to move-box A moves up, while box B moves down.

.

Free-body diagram of box A:

  

If box A remains at rest, then

  $\begin{array}{l}{\displaystyle \sum F_{\text{ext,Ay}}=T+n-w_{\text{A}}}=m_{\text{A}}{a}_{\text{y}}=0\\ =>n=w_{\text{A}}-T\end{array}$

If box B weighs more than box A, then we would expect box B to fall down and lift box A off

The table.

Since box A is no longer touching the table, the normal goes to $0$.

Conclusion:

The force that table exerts on box A if box B weighs $90\text{N}$ is $=0\text{}\text{N}$, since box A is no longer touching the table, the normal goes to $0$.