Question

lesson 4 newtons third law
lesson practice

1. identify each force acting on the ball and its interaction pair in the figure below
2. look again at the figure of the hand holding the ball. imagine lowering the ball at increasing speed. draw separate free - body diagrams for the forces acting on the ball and for each set of interaction pairs.
3. a block hangs from the ceiling by a massless rope. a second block is attached to the first block and hangs below it on another piece of massless rope. if each of the two blocks has a mass of 5.0 kg, what is the tension in each rope?
4. a block hangs from the ceiling by a massless rope. a 3.0 - kg block is attached to the first block and hangs below it on another piece of massless rope. the tension in the top rope is 63.0 n. find the tension in the bottom rope and the mass of the top block.
5. a curtain prevents two tug - of - war teams from seeing each other. one team ties its end of the rope to a tree. if the other team pulls with a 500 - n force, what is the tension in the rope? explain.

Explanation:

Step1: Analyze forces on ball in hand - case 1

The force of gravity $F_g = mg$ acts downwards on the ball, where $m$ is the mass of the ball and $g = 9.8\ m/s^{2}$. The interaction - pair is the gravitational force the ball exerts on the Earth. The normal force $N$ exerted by the hand on the ball acts upwards. The interaction - pair is the force the ball exerts on the hand.

Step2: Analyze forces on ball being lowered

The force of gravity $F_g=mg$ still acts downwards. Since the ball is accelerating downwards, $F_g>N$, where $N$ is the normal force exerted by the hand on the ball. The interaction - pairs remain the same as in Step 1.

Step3: Solve for tension in ropes for two - block system (question 3)

Let $m_1 = m_2=5.0\ kg$. For the lower block, the net - force equation is $F_{net}=m_2a$. In equilibrium ($a = 0$), $T_2=m_2g$, where $T_2$ is the tension in the lower rope. So $T_2=5.0\times9.8 = 49\ N$. For the upper block, $T_1=T_2 + m_1g$, so $T_1=49+5.0\times9.8=98\ N$.

Step4: Solve for tension and mass in two - block system (question 4)

Let the mass of the upper block be $m_1$ and the mass of the lower block $m_2 = 3.0\ kg$. The tension in the lower rope $T_2=m_2g=3.0\times9.8 = 29.4\ N$. The tension in the upper rope $T_1=T_2 + m_1g$. We know $T_1 = 63.0\ N$, so $m_1=\frac{T_1 - T_2}{g}=\frac{63.0 - 29.4}{9.8}=\frac{33.6}{9.8}=3.43\ kg$.

Step5: Solve for tension in tug - of - war rope (question 5)

In a tug - of - war situation, when one team ties the rope to a tree and the other team pulls with a force $F = 500\ N$, the rope is in equilibrium. According to Newton's third law, the tension in the rope $T$ is equal to the force applied by the pulling team. So $T = 500\ N$.

Answer:

1. Force of gravity on ball (downwards), interaction - pair: ball's gravitational pull on Earth; Normal force by hand on ball (upwards), interaction - pair: ball's force on hand.
2. Draw free - body diagrams with gravity downwards and normal force upwards on the ball, with gravity being greater than the normal force as the ball is accelerating downwards. Interaction pairs: ball - Earth gravitational pair and ball - hand normal - force pair.
3. Tension in lower rope: $49\ N$, Tension in upper rope: $98\ N$.
4. Tension in bottom rope: $29.4\ N$, Mass of top block: $3.43\ kg$.
5. Tension in the rope: $500\ N$.