Solutions to the practice exercises on tension and connected particles

This entry gives worked solutions to the practice exercises posed in the companion article Tension and connected particles. The main goals are to reinforce the use of Free-body diagrams, to apply Newton's second law consistently, and to use string-length constraints correctly.

Throughout, take

$\displaystyle g=9.81\,\mathrm{m/s^2}$

unless a symbolic result is requested.

Problem 1

A $6\,\mathrm{kg}$ mass hangs at rest from a vertical rope. Find the tension.

Since the mass is at rest, $a=0$ . Using upward as positive,

$\displaystyle T-mg=0.$

Thus

$\displaystyle T=mg=(6)(9.81)=58.86\,\mathrm{N}.$

Answer:

$\displaystyle \boxed{T=58.9\,\mathrm{N}}.$

Problem 2

The same mass accelerates upward at $2\,\mathrm{m/s^2}$ . Find the tension.

Using

$\displaystyle T-mg=ma,$

we obtain

$\displaystyle T=m(g+a)=6(9.81+2)=70.86\,\mathrm{N}.$

Answer:

$\displaystyle \boxed{T=70.9\,\mathrm{N}}.$

Problem 3

The same mass accelerates downward at $2\,\mathrm{m/s^2}$ . Find the tension.

If upward is positive, then $a=-2\,\mathrm{m/s^2}$ , so

$\displaystyle T-mg=m(-2).$

Therefore

$\displaystyle T=m(g-2)=6(9.81-2)=46.86\,\mathrm{N}.$

Answer:

$\displaystyle \boxed{T=46.9\,\mathrm{N}}.$

Problem 4

Two blocks of $2$ and $3\,\mathrm{kg}$ are pulled on a smooth table by $20\,\mathrm{N}$ . Find the acceleration and the tension.

Let $m_1=2\,\mathrm{kg}$ and $m_2=3\,\mathrm{kg}$ . Treating the two blocks as one system,

$\displaystyle a=\frac{F}{m_1+m_2}=\frac{20}{5}=4.0\,\mathrm{m/s^2}.$

Now isolate $m_1$

$\displaystyle T=m_1a=(2)(4.0)=8.0\,\mathrm{N}.$

Answer:

$\displaystyle \boxed{a=4.0\,\mathrm{m/s^2}},\qquad \boxed{T=8.0\,\mathrm{N}}.$

Problem 5

A $5\,\mathrm{kg}$ block on a smooth table is connected to a hanging $2\,\mathrm{kg}$ mass. Find $a$ and $T$ .

Let $m_1=5\,\mathrm{kg}$ on the table and $m_2=2\,\mathrm{kg}$ hanging. The equations are

$\displaystyle T=m_1a,$

$\displaystyle m_2g-T=m_2a.$

Adding,

$\displaystyle m_2g=(m_1+m_2)a,$

$\displaystyle a=\frac{2(9.81)}{7}=2.80\,\mathrm{m/s^2}.$

Then

$\displaystyle T=m_1a=(5)(2.80)=14.0\,\mathrm{N}.$

Answer:

$\displaystyle \boxed{a=2.80\,\mathrm{m/s^2}},\qquad \boxed{T=14.0\,\mathrm{N}}.$

Problem 6

Solve the ideal Atwood machine for $m_1=4\,\mathrm{kg}$ and $m_2=7\,\mathrm{kg}$ .

Because $m_2>m_1$ , the $7\,\mathrm{kg}$ mass moves downward. The acceleration magnitude is

$\displaystyle a=\frac{m_2-m_1}{m_1+m_2}g =\frac{7-4}{11}9.81 =2.68\,\mathrm{m/s^2}.$

The tension is

$\displaystyle T=\frac{2m_1m_2}{m_1+m_2}g =\frac{2(4)(7)}{11}(9.81) =49.9\,\mathrm{N}.$

Answer:

$\displaystyle \boxed{a=2.68\,\mathrm{m/s^2}},\qquad \boxed{T=49.9\,\mathrm{N}}.$

Problem 7

Derive the general Atwood acceleration without first solving for $T$ .

For the lighter mass $m_1$ moving upward,

$\displaystyle T-m_1g=m_1a.$

For the heavier mass $m_2$

moving downward,

$\displaystyle m_2g-T=m_2a.$

Add the equations so that $T$

cancels:

$\displaystyle m_2g-m_1g=(m_1+m_2)a.$

Therefore

$\displaystyle \boxed{a=\frac{m_2-m_1}{m_1+m_2}g}.$

Problem 8

A $4\,\mathrm{kg}$ block lies on a smooth $25^\circ$ incline and is connected to a $3\,\mathrm{kg}$ hanging mass. Determine the direction of motion.

Compare the driving weights along the string. For the incline block,

$\displaystyle m_1g\sin25^\circ=4(9.81)\sin25^\circ\approx16.58\,\mathrm{N}.$

For the hanging block,

$\displaystyle m_2g=3(9.81)=29.43\,\mathrm{N}.$

Since

$\displaystyle m_2g>m_1g\sin25^\circ,$

the hanging mass pulls downward and the incline block moves upward.

Answer: the $3\,\mathrm{kg}$ mass moves downward and the $4\,\mathrm{kg}$ block moves up the incline.

Problem 9

Repeat the preceding problem if $\mu_k=0.15$ .

If the block on the incline moves upward, the kinetic friction acts down the plane. The resisting force on the incline side becomes

$\displaystyle m_1g\sin25^\circ+\mu_k m_1g\cos25^\circ.$

Numerically,

$\displaystyle m_1g\sin25^\circ\approx16.58\,\mathrm{N},$

$\displaystyle \mu_k m_1g\cos25^\circ =0.15(4)(9.81)\cos25^\circ \approx5.34\,\mathrm{N}.$

Total resistance on the incline side:

$\displaystyle 16.58+5.34=21.92\,\mathrm{N}.$

Since

$\displaystyle 29.43>21.92,$

the hanging mass still moves downward.

If one also wants the acceleration,

$\displaystyle a=\frac{m_2g-m_1g\sin25^\circ-\mu_k m_1g\cos25^\circ}{m_1+m_2} =\frac{29.43-21.92}{7} =1.07\,\mathrm{m/s^2}.$

Answer: the direction is unchanged; the $3\,\mathrm{kg}$ mass still moves downward. The acceleration is

$\displaystyle \boxed{a\approx1.07\,\mathrm{m/s^2}}.$

Problem 10

Three masses in series have masses $1,2,4\,\mathrm{kg}$ and are pulled by $28\,\mathrm{N}$ . Find both string tensions.

The total mass is

$\displaystyle 1+2+4=7\,\mathrm{kg},$

$\displaystyle a=\frac{28}{7}=4.0\,\mathrm{m/s^2}.$

For the first string,

$\displaystyle T_1=m_1a=(1)(4.0)=4.0\,\mathrm{N}.$

For the second string, treat $m_1+m_2$

as a subsystem:

$\displaystyle T_2=(m_1+m_2)a=(1+2)(4.0)=12.0\,\mathrm{N}.$

Answer:

$\displaystyle \boxed{T_1=4.0\,\mathrm{N}},\qquad \boxed{T_2=12.0\,\mathrm{N}}.$

Problem 11

A uniform rope of length $L$ and mass $M$ hangs vertically supporting a mass $m$ . Find the tension at the bottom and top of the rope.

At the bottom of the rope, the tension supports only the attached mass:

$\displaystyle T_{\rm bottom}=mg.$

At the top of the rope, the tension supports the attached mass plus the full rope mass:

$\displaystyle T_{\rm top}=(m+M)g.$

Answer:

$\displaystyle \boxed{T_{\rm bottom}=mg},\qquad \boxed{T_{\rm top}=(m+M)g}.$

Problem 12

Show that a massless pulley implies equal tension if bearing friction is neglected.

Let the tensions on the two sides be $T_1$ and $T_2$ , and let the pulley radius be $R$ . The torque equation is

$\displaystyle (T_2-T_1)R=I\alpha.$

For a massless pulley, $I=0$

. If bearing friction is neglected, then

$\displaystyle (T_2-T_1)R=0,$

$\displaystyle T_2=T_1.$

Answer:

$\displaystyle \boxed{T_1=T_2}$

for a massless frictionless pulley.

Problem 13

For a massive pulley with $I=\frac12MR^2$ , derive $T_2-T_1$ in terms of $M$ and $a$ .

The pulley torque equation gives

$\displaystyle (T_2-T_1)R=I\alpha.$

Because the string does not slip,

$\displaystyle a=\alpha R, \qquad \alpha=\frac{a}{R}.$

Hence

$\displaystyle T_2-T_1=\frac{I}{R^2}a.$

Substitute

$\displaystyle I=\frac12MR^2:$

$\displaystyle T_2-T_1=\frac{\frac12MR^2}{R^2}a=\frac12Ma.$

Answer:

$\displaystyle \boxed{T_2-T_1=\frac12Ma}.$

Problem 14

Explain physically what must happen if an algebraic solution gives $T<0$ .

A flexible string can pull but cannot push. Therefore a negative tension is not physically admissible for the assumed taut-string configuration. The interpretation is that the string would go slack, so the original constraint model no longer applies. One must reformulate the motion without imposing a taut inextensible string over that interval.

Answer: a negative tension means the assumed taut-string model has failed; the string goes slack.

General remarks

Several themes recur in these solutions:

Always draw the free-body diagram before writing equations.
Use one coordinate direction for each body that aligns with its motion.
If the same string connects two bodies, the magnitude of their accelerations is often fixed by the constraint.
Equal tension requires an ideal continuous string and an ideal pulley.
For several masses in series, different strings generally carry different tensions.

Source and licensing note

These solutions are original PhysicsLibrary content written as a companion to the article Tension and connected particles. The problem statements are the same as those posed in the practice section of that companion entry. The figures in this article were generated specifically for this solution set.

Bibliography

1: PhysicsLibrary, Tension and connected particles.
2: Physics LibreTexts, connected-particle and Atwood-machine examples. Physics LibreTexts

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