Question

Four weights, of masses \(m_{1}=6.50 \mathrm{~kg}_{3}\) \(m_{2}=3.80 \mathrm{~kg}, m_{3}=10.70 \mathrm{~kg},\) and \(m_{4}=\)

Short answer

Explain your answer. Answer: No, the system cannot be brought into equilibrium by adjusting the mass of M only. For the given configuration of weights, the calculated value of M is -2.4 kg, which is not physically valid as mass cannot be negative. It indicates that the configuration of the weights needs to be rearranged, or their values changed, to achieve equilibrium.

Step-by-step solution

Analyze the horizontal forces

: As the strings and pulleys are frictionless, the tension in each of the horizontal strings will be equal and opposite. Label the tensions as \(T_1\) and \(T_2\). We can write the following equation: \(T_1=T_2\)

Analyze the vertical forces

: There are vertical forces acting upon the system; let's break them down for each mass: - For \(m_1\): The vertical force is \(m_1g\), where \(g\) is the acceleration due to gravity. - For \(m_2\): The vertical force is \(m_2g\). - For \(m_3\): The vertical force is \(m_3g\). - For \(m_4\): The vertical force is \(m_4g\). - For M: The vertical force is \(Mg\). Now, apply the condition of equilibrium for vertical forces: \(m_1g - m_2g - m_3g + m_4g - Mg = 0\)

Solve for M

: Now, plug in the given values for the masses and the acceleration due to gravity (\(g \approx 9.81 \mathrm{m/s^2}\)), and solve for \(M\): \((6.50)9.81 - (3.80)9.81 - (10.70)9.81 + (5.60)9.81 - M(9.81) = 0\) Solve for \(M\): \(M = \dfrac{(6.50 - 3.80 - 10.70 + 5.60)9.81}{9.81}\) \(M = \dfrac{(-2.4)9.81}{9.81}\) \(M = -2.4 \,\mathrm{kg}\)

Consider the result

: The result of \(M = -2.4 \,\mathrm{kg}\) indicates that, for the given configuration, it's not possible to maintain equilibrium with a physically valid mass for M (mass cannot be negative). It means the configuration of the weights needs to be rearranged, or their values changed, to achieve equilibrium.

Key concepts

Forces and Motion

Understanding the interplay of forces and motion is crucial in physics, especially when analyzing static or dynamic systems. In equilibrium scenarios, the net force acting on an object must be zero, meaning all the individual forces balance each other out.

In the case of stationary objects, this state is known as static equilibrium. When multiple weights and strings are involved, as in the textbook exercise, the principle of equilibrium applies to both the horizontal and vertical components of the forces. For horizontal forces, they must counterbalance each other so that their vector sum equals zero. This is why we set the tension in the strings to be equal and opposite.

• If an object is in equilibrium, the sum of the forces in any direction must be zero.
• For forces in motion (dynamic equilibrium), despite moving, the object's net force is still zero, resulting in a constant velocity.
• Understanding vector addition is vital for resolving forces into their horizontal and vertical components.
• The equilibrium condition can help predict an object's motion or lack thereof, given the applied forces.

Tension in Strings

Tension is the force conducted through a string, rope, or wire when it is pulled tight by forces acting from opposite ends. In the context of physics problems, such as the textbook exercise, tension in strings plays a major role in systems involving pulleys and hanging masses.

In a perfect, ideal scenario (i.e., massless and frictionless strings and pulleys), the tension throughout the string is constant. This assumption simplifies the analysis considerably:

• Tension forces point away from the object applying the force and are equal and opposite to the forces causing the tension.
• In a system of interconnected strings and pulleys, the tension must be analyzed at each point to understand the equilibrium state.
• The property that the tension is the same at all points along the string only holds if the pulleys are frictionless and the string is massless.

Applying these principles to the given problem helps to isolate variables and solve for unknowns like mass or tension, considering the gravitational force acting on the weights.

Acceleration due to Gravity

The acceleration due to gravity, represented by the symbol 'g', is a constant value used to quantify the effect of Earth's gravitational pull on objects. It has a standard value of approximately 9.81 m/s² near the surface of the Earth.

When dealing with weights and forces in physics problems, the force exerted by gravity on an object with mass 'm' is calculated as the product of the mass and the acceleration due to gravity (F = mg). This force is also known as weight.

• Every object near the Earth's surface experiences this acceleration equally, provided air resistance is neglected.
• In textbooks solutions, the variable 'g' allows us to calculate the gravitational force regardless of the weight's mass.
• The direction of gravitational force is always towards the center of the Earth, or downward in most diagrams.

In the given exercise, gravity's role is essential because it directly affects the tension in the strings and ultimately determines if the system can be in equilibrium. By using the concept of acceleration due to gravity, we can calculate the individual weights of objects and their contribution to the overall force balance.