Worked examples · Sheet 2

Bodies in equilibrium hanging from strings

Theory used: Equilibrium of a point particle Tension of a string Tension along an ideal string Ideal pulley Principle of superposition of forces Free-body diagram

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

A lamp of mass $m_{1} = 3.0 kg$ hangs from the ceiling by a vertical string. From the lamp hangs, by a second vertical string, a second lamp of mass $m_{2} = 2.0 kg$ . The strings are ideal and the system is at rest. Compute the tension of the two strings.

lower lamp: T_{2} - m_{2} g = 0 ⟹ T_{2} = m_{2} g

upper lamp: T_{1} - T_{2} - m_{1} g = 0 ⟹ T_{1} = (m_{1} + m_{2}) g

T_{2} = 2.0 kg \cdot 9.81 m/ s^{2} = 19.6 N, T_{1} = 5.0 kg \cdot 9.81 m/ s^{2} = 49.1 N

The equilibrium condition is written for each lamp, with the vertical axis pointing upward. On the lower lamp act its weight and the tension of the lower string, therefore $T_{2} = m_{2} g$ . On the upper lamp act its weight, the tension $T_{1}$ of the upper string, upward, and the tension $T_{2}$ of the lower string, which pulls downward.

The lower string is ideal, so it pulls the upper lamp with the same tension $T_{2}$ with which it supports the lower one. It follows that the upper string supports the weight of both lamps, $49.1 N$ . The same result is obtained by considering the two lamps and the lower string as a single body of mass $5.0 kg$ , on which the tension $T_{2}$ is an internal force.

Exercise 2

A picture of mass $m = 6.0 kg$ hangs from a nail by means of an ideal string, fixed to the two sides of the frame, that passes over the nail without friction; the two segments of string are symmetric and each forms an angle of $6 0^{\circ}$ with the horizontal. Compute the tension of the string. Then compute the tension if the string is shortened so that the two segments form an angle of $1 0^{\circ}$ with the horizontal.

{T cos θ - T cos θ = 0 2 T sin θ - m g = 0 ⟹ T = \frac{m g}{2 sin θ}

θ = 6 0^{\circ} : T = \frac{6.0 kg \cdot 9.81 m/ s ^{2}}{2 \cdot 0.866} = \frac{58.9 N}{1.732} = 34.0 N

θ = 1 0^{\circ} : T = \frac{58.9 N}{2 \cdot 0.1736} = 170 N

The picture is treated as a point particle. On it act its weight and the tensions of the two segments of string, directed along the segments toward the nail, where the lines of action of the three forces meet. Since the string is ideal and passes over the nail without friction, the two segments have the same tension $T$ . The horizontal components of the two tensions have the same magnitude $T cos θ$ and opposite senses, therefore the first equation is satisfied for every value of $T$ . The vertical components both point upward, and their sum $2 T sin θ$ balances the weight.

With the segments at $6 0^{\circ}$ the tension is $34.0 N$ , a little more than half the weight. With the segments at $1 0^{\circ}$ the sine is much smaller, therefore the tension rises to $170 N$ , almost three times the weight. In the limiting case $θ \to 0$ the tension would tend to infinity, so a string stretched horizontally cannot support a weight.

Exercise 3

A traffic light of mass $m = 10 kg$ is supported by two ideal cables fixed to two poles. The left cable forms an angle of $3 0^{\circ}$ with the horizontal, the right one an angle of $6 0^{\circ}$ . Compute the tensions $T_{1}$ (left cable) and $T_{2}$ (right cable).

{T_{2} cos 6 0^{\circ} - T_{1} cos 3 0^{\circ} = 0 T_{1} sin 3 0^{\circ} + T_{2} sin 6 0^{\circ} - m g = 0

T_{2} = T_{1} \frac{cos 3 0 ^{\circ}}{cos 6 0 ^{\circ}} = 3 T_{1}

T_{1} \cdot \frac{1}{2} + 3 T_{1} \cdot \frac{3}{2} - m g = 0 ⟹ 2 T_{1} = m g ⟹ T_{1} = \frac{m g}{2}

T_{1} = \frac{10 kg \cdot 9.81 m/ s ^{2}}{2} = 49.1 N, T_{2} = 3 \cdot 49.05 N = 85.0 N

Forces on the traffic light, with the arrows to scale (

25 N

per unit): the horizontal components of

T_{1}

and

T_{2}

are equal and opposite

On the traffic light act its weight and the tensions of the two cables, each directed along its own cable toward the pole. The $x$ axis is chosen horizontal and pointing right and the $y$ axis vertical and pointing upward. The tension $T_{1}$ has components $(- T_{1} cos 3 0^{\circ}, T_{1} sin 3 0^{\circ})$ , the tension $T_{2}$ has components $(T_{2} cos 6 0^{\circ}, T_{2} sin 6 0^{\circ})$ . The traffic light is at rest, therefore the net force is zero and the two components give a linear system in the unknowns $T_{1}$ and $T_{2}$ .

From the first equation one obtains $T_{2}$ in terms of $T_{1}$ ; substituting into the second one obtains $T_{1} = \frac{m g}{2} = 49.1 N$ and therefore $T_{2} = 85.0 N$ . The cable closer to the vertical bears the greater tension. As a check, the two cables are perpendicular, because $3 0^{\circ} + 6 0^{\circ} = 9 0^{\circ}$ , so the tensions are the projections of the weight onto the two directions: $T_{1} = m g sin 3 0^{\circ}$ and $T_{2} = m g cos 3 0^{\circ}$ , in agreement with the values found.

Exercise 4

A load of mass $m = 20 kg$ hangs from a movable pulley. The string that supports the movable pulley has one end fixed to the ceiling, passes under the movable pulley, then over a pulley fixed to the ceiling, and the other end is pulled downward by a worker. Strings and pulleys are ideal and the segments of string that support the movable pulley are vertical. Compute the force the worker must exert to lift the load at constant velocity and the force needed to lift it with acceleration $a = 0.50 m/ s^{2}$ upward.

2 T - m g = m a ⟹ T = \frac{m ( g + a )}{2}

a = 0 : T = \frac{20 kg \cdot 9.81 m/ s ^{2}}{2} = 98.1 N

a = 0.50 m/ s^{2} : T = \frac{20 kg \cdot 10.31 m/ s ^{2}}{2} = 103 N

The body considered is the load together with the movable pulley, which has negligible mass. On it act the weight of the load and two upward forces, one for each segment of string passing under the pulley. The string is ideal and the pulleys are ideal, therefore the tension is the same $T$ in every segment, including the one pulled by the worker; so the force of the worker is exactly $T$ .

At constant velocity the acceleration is zero and the second law reduces to the equilibrium $2 T = m g$ , therefore the worker exerts half the weight, $98.1 N$ . With upward acceleration the net force must equal $ma$ , so the force rises to $103 N$ . The fixed pulley only changes the direction of the force, allowing the worker to pull downward.