Question

A hanging mass, $M_1=0.50\mathrm{kg}$, is attached by a light string that runs over a frictionless pulley to the front of a mass $M_2=1.50\mathrm{kg}$ that is initially at rest on a frictionless table. A third mass $M_3=2.50\mathrm{kg}$, which is also initially at rest on a frictionless table, is attached to the back of $M_2$ by a light string. a) Find the magnitude of the acceleration, $a,$ of mass $M_3$ b) Find the tension in the string between masses $M_1$ and $M_2$.

Short answer

Answer: The magnitude of the acceleration of mass M_3 is approximately 1.089 m/s^2, and the tension in the string between masses M_1 and M_2 is approximately 5.445 N.

Step-by-step solution

Analyze the forces acting on each mass

For each mass, we identify the forces acting on them: - Mass $M_1$: the gravitational force ($M_{1}g$) acts downwards, and the tension force $T_1$ due to the string acts upwards. - Mass $M_2$: the tension force $T_1$ due to the string acts horizontally to the right, and the tension force $T_2$ linking it to mass $M_3$ acts to the left. - Mass $M_3$: the horizontal tension force $T_2$ acts to the right.

Apply Newton's second law for each mass

For each mass, we can write equations corresponding to Newton's second law for appropriate directions (vertical for $M_1$ and horizontal for $M_2$ and $M_3$): - Mass $M_1$: $T_1-M_{1}g=M_{1}a$ (upwards direction is positive) - Mass $M_2$: $T_1-T_2=M_{2}a$ (right direction is positive) - Mass $M_3$: $T_2=M_{3}a$ (right direction is positive)

Calculate the acceleration, $a$

Now we can solve the system of equations formed by Steps 2. First, add the equations for mass $M_2$ and mass $M_3$ to eliminate the tension force $T_2$: $T_1=(M_2+M_3)a$ Now, rearrange the equation for mass $M_1$: $T_1=M_{1}g+M_{1}a$ Now, substitute the expression for $T_1$ from the (the sum of mass $M_2$ and $M_3$ equations) into the (mass $M_1$ equation): $(M_2+M_3)a=M_{1}g+M_{1}a$ Now solve for $a$: $a={\displaystyle \frac{M_{1}g}{(M_2+M_3)+M_1}}$ Plug in the given values for the masses and the acceleration due to gravity $g=9.8\mathrm{m}/{\mathrm{s}}^2$: $a={\displaystyle \frac{0.50\times 9.8}{(1.50+2.50)+0.50}}={\displaystyle \frac{4.9}{4.5}}\approx 1.089\mathrm{m}/{\mathrm{s}}^2$ This is the magnitude of the acceleration of mass $M_3$.

Calculate the tension force between masses $M_1$ and $M_2$, $T_1$

To find the tension force $T_1$, plug in the found acceleration value and the given mass value for $M_1$ into any of the previous equations, for example, the equation for mass $M_2$: $T_1=M_{1}g+M_{1}a=0.50\times 9.8+0.50\times 1.089\approx 5.445\mathrm{N}$ Now we have found the tension force between masses $M_1$ and $M_2$.

Results

The magnitude of the acceleration of mass $M_3$ is $1.089\mathrm{m}/{\mathrm{s}}^2$, and the tension in the string between masses $M_1$ and $M_2$ is $5.445\mathrm{N}$.

Key concepts

Acceleration Calculation

Understanding acceleration in physics is crucial when examining the motion of objects. Acceleration is the rate at which an object changes its velocity. It is a vector quantity, meaning it has both magnitude and direction. According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass.

The formula derived from Newton's second law for calculating acceleration is:$$a=\frac{F_{net}}{m}$$where $a$ is acceleration, $F_{net}$ is the net force, and $m$ is the mass of the object. In the context of the exercise provided, we're dealing with a pulley system where the gravitational force acting on mass $M_1$ influences the acceleration of the other masses. The equation to find acceleration becomes more specific:$$a=\frac{M_{1}g}{M_{total}+M_1}$$where $M_{total}$ is the sum of the masses $M_2$ and $M_3$. By substituting the values for the masses and the acceleration due to gravity $g$, students can calculate the acceleration. It's essential to substitute the correct values and perform the arithmetic carefully to obtain the correct answer.

Tension Force

Tension is a force that is transmitted through a string, cable, or wire when it is pulled tight by forces acting from opposite ends. In physics problems, tension is often considered as a force that tries to restore the material to its original length. In the frictionless pulley system described in the exercise, the tension force comes into play because the string transmits forces between the different masses.

In our case, the tension force can be calculated using the equations derived from Newton's second law. For example, the tension force $T_1$ in the string between masses $M_1$ and $M_2$ has an effect on both masses. Since the system is frictionless, and no other external forces are acting on the horizontal surface, the only forces in the horizontal direction are the tension forces.

The exercise guides students through using the correct equations to find the tension force:$$T_1-M_{1}g=M_{1}a$$ and $$T_1=M_{2}a+T_2$$After solving for the acceleration, these equations can be used to determine tension. It's vital for students to understand that the value for acceleration must be consistent across all parts of the system for these calculations to work correctly.

Frictionless Pulley System

A pulley system consists of one or more wheels with a rope or cable running along its groove. It is used to change the direction of a force and can also multiply the force, making it easier to lift weights. However, when we introduce the idea of a frictionless pulley system in physics, we are simplifying the real-world scenario to eliminate the complexities introduced by friction.

In such a system, the pulleys do not resist the motion of the rope, which means all the tension in the rope is conserved across the system. This lack of resistance simplifies calculations because it allows us to ignore the energy loss that would be caused by friction in a real-world application.

Students should appreciate that in a frictionless pulley system, the masses are connected via a rope not subject to any wear or tear, and the pulley does not add any additional force to the system. This assumption is critical for solving problems using Newton's second law. By making these simplified assumptions, students can focus on the fundamental concepts of force and acceleration without getting bogged down by more complex aspects of real-world physics.