Question

$FIGUREEX7.16$shows two $1.0kg$blocks connected by a rope. A second rope hangs beneath the lower block. Both ropes have a mass of $250g$. The entire assembly is accelerated upward at $3.0m/s^2$by force $\overrightarrow{F}$

a. What is F?

b. What is the tension at the top end of rope $1$?

c. What is the tension at the bottom end of rope $1$?

d. What is the tension at the top end of rope $2$?

Short answer

(a) The force F is $32N$

(b) The tension at the upper part of the rope $1$is $19.2N$

(c) The tension at the lower part of the rope $1$is $16N$

(d) The tension at the upper part of the rope $2$is $3.2N$

Step-by-step solution

Given information (part a)

Mass of rope $1=0.25kg$

Mass of rope $2=0.25kg$

Mass of $B=1kg$

Acceleration of the whole system$=3m/s^2$

Explanation (part a)

Consider the motion of the system:

For this, consider different forces acting in their directions as

Weight on the system (downward), W = weight of B + Weight of rope $1$+ Weight of rope $2$
localid="1650277077562" $$\begin{gathered} W=m_{1}g+m_{2}g+mg+mg=1(9.8)+1(9.8)+0.25(9.8)+0.25(9.8) \\ W=24.5\mathrm{N} \end{gathered}$$

The force on the system (upward), F = Force on A + Force on B + Force on rope $1$+ Force on the rope $2$
localid="1650277035728" $$\begin{gathered} m_{1}a+m_{2}a+ma+ma=1\left(3\right)+1\left(3\right)+0.25\left(3\right)+0.25\left(3\right) \\ {F}_{net}=7.5\mathrm{N} \end{gathered}$$

Substituting the values

localid="1650277005242" $$\begin{gathered} F-W=F_{net} \\ F=W+F_{net} \\ F=24.5+7.5 \\ =32\mathrm{N} \end{gathered}$$

Given information (part b)

Mass of rope $1=0.25kg$

Mass of rope $2=0.25kg$

Mass of $B=1kg$

Acceleration of the whole system$=3m/s^2$

Explanation (part b)

Consider the motion of the system just below block A.

For this, consider different forces acting in their directions as

Weight on the system (downward), W = weight of B + Weight of rope $1$+ Weight of rope $2$

localid="1649006557408" $$\begin{gathered} m_{2}g+mg+mg=1(9.8)+0.25(9.8)+0.25(9.8) \\ =14.7\mathrm{N} \end{gathered}$$

The force on the system (upward), F = Force on A + Force onB + Force on rope $1$+ Force on rope $2$

localid="1649006624893" $$\begin{gathered} m_{2}a+ma+ma=1\left(3\right)+0.25\left(3\right)+0.25\left(3\right) \\ =4.5\mathrm{N} \\ \mathrm{F}+\mathrm{W}=\mathrm{T} \\ \text{Substitutingthevaluesofgivendata} \\ \Rightarrow \mathrm{T}=(14.7+4.5) \\ =19.2\mathrm{N} \end{gathered}$$

Given information (part c)

Mass of rope $1=0.25kg$

Mass of rope $2=0.25kg$

Mass of $B=1kg$

Acceleration of the whole system$=3m/s^2$

Explanation (part c)

Consider the motion of the system:

For this, consider different forces acting in their directions as

Weight on the system (downward), W = weight of B + Weight of rope 1 + Weight of rope$2$

$$\begin{gathered} m_{2}g+mg=1(9.8)+0.25(9.8) \\ =12.25\mathrm{N} \end{gathered}$$

The force on the system (upward), F = Force on A + Force on B + Force on rope $1$+ Force on the rope $2$

$$\begin{gathered} m_{2}a+ma=1\left(3\right)+0.25\left(3\right) \\ =3.75\mathrm{N} \\ \mathrm{F}+\mathrm{W}=\mathrm{T} \\ \text{Substitutingthevaluesofgivendata} \\ \mathrm{T}=(12.25+3.75) \\ =16\mathrm{N} \end{gathered}$$

Given information (part d)

Mass of rope $1=0.25kg$

Mass of rope $2=0.25kg$

Mass of $B=1kg$

Acceleration of the whole system$=3m/s^2$

Explanation (part d)

Consider different forces acting with their directions as weight (downward),

W= weight of the rope 2

$$\begin{gathered} mg=\left(0.25\right)\left(9.8\right) \\ =2.45N\left(downward\right) \end{gathered}$$

The force (upward), F= Force of rope 2

$$\begin{gathered} m_{2}a+ma=0.25\left(3\right) \\ =0.75N \end{gathered}$$

The equation of motion is

$F+W=T\dots \dots \left(1\right)$

Substituting the value of the given data in equation$\left(1\right)$

$$\begin{gathered} T=0.75+2.45 \\ =3.2N \end{gathered}$$

Tension T is$3.2N$