Question

A mobile is constructed from a metal bar and two wooden blocks, as shown in the figure. The metal bar has a mass of \(1.0 \mathrm{~kg}\) and is \(10 . \mathrm{cm}\) long. The metal bar has a \(3.0-\mathrm{kg}\) wooden block hanging from the left end and a string tied to it at a distance of \(3.0 \mathrm{~cm}\) from the left end. What mass should the wooden block hanging from the right end of the bar have to keep the bar level? a) \(0.70 \mathrm{~kg}\) b) \(0.80 \mathrm{~kg}\) c) \(0.90 \mathrm{~kg}\) d) \(1.0 \mathrm{~kg}\) e) \(1.3 \mathrm{~kg}\) f) \(3.0 \mathrm{~kg}\) g) \(7.0 \mathrm{~kg}\)

Short answer

(rounded to 3 decimal places) A. 0.198 kg B. 0.2 kg C. 0.25 kg D. 0.3 kg The answer is A. 0.198 kg.

Step-by-step solution

Find the force of the 3-kg wooden block

We need to calculate the force exerted by the 3-kg wooden block. This force is the gravitational force, \(F_1 = m_1\times g\), where \(m_1 = 3.0 kg\) and \(g = 9.81 m/s^2\). \(F_1 = 3.0 kg \times 9.81 m/s^2 = 29.43 N\)

Calculate the torque exerted by the 3-kg wooden block

Now we need to calculate the torque exerted by the 3-kg wooden block at its position. The equation for torque is \(\tau_1 = r_1F_1sin(\theta_1)\) When the bar is hanging perpendicular the the distance between the torque, the angle is 90 degrees. So the value of \(sin(\theta_1) = 1\) and the distance \(r_1 = 3.0 cm\) from the pivot point. \(\tau_1 = (3.0 cm \times 0.01)(29.43 N)(1) = 0.8829 Nm\)

Calculate the force and torque exerted by the metal bar

The force exerted by the metal bar is the gravitational force, \(F_2 = m_2\times g\), where \(m_2 = 1.0 kg\) and g is still \(9.81 m/s^2\). \(F_2 = 1.0 kg \times 9.81 m/s^2 = 9.81 N\) The distance between the center of mass of the bar and the pivot point is half the length, which is \(5 cm = 0.05 m\). Therefore, the torque exerted by the metal bar is given by \(\tau_2 = r_2F_2sin(\theta_2)\) \(\tau_2 = (0.05 m)(9.81 N)(1) = 0.4905 Nm\)

Calculate the total torque on the left side

We need to find the total torque exerted on the left side of the mobile. This will help us find the torque needed on the right side to balance the bar. The torques are additive. \(\tau_{total}= \tau_{1} + \tau_{2} = 0.8829 Nm + 0.4905 Nm = 1.3734 Nm\)

Calculate the mass of the wooden block on the right side

To balance the bar, the torque on the right side should be equal to the total torque on the left side. We have to find the mass \(m_3\) of the wooden block on the right side. Using the torque equation again, we get: \(\tau_{total} = r_3F_3sin(\theta_3)\) As the angle is 90 degrees, the sin value will be 1. The distance is given by \(r_3 = 7 cm\). So the expression becomes: \(1.3734 Nm = 0.07 m \times (m_3\times g)(1)\) To find \(m_3\), isolate \(m_3\) in the equation: \(m_3 = \frac{1.3734 Nm}{0.07 m\times g} = \frac{1.3734 Nm}{0.07 m\times 9.81 m/s^2} \approx 0.198 kg\) Hence, the mass of the wooden block hanging from the right end needs to be approximately 0.198 kg to keep the bar level. Since there is no option available in the choice near 0.198 kg, it might be a mistake in the question or the given options.

Key concepts

Gravitational Force

Gravitational force is a fundamental interaction that exists between all masses. It is the force responsible for the mutual attraction between the Earth and objects, causing them to have weight. This force is essential when solving problems related to objects at rest or in motion near the Earth’s surface. In the calculations of the given exercise, gravitational force is represented by the acceleration due to gravity, denoted by the symbol 'g', which has a standard value of approximately \( 9.81 \, \text{m/s}^2 \).

When we calculate the force exerted by an object due to gravity, we use the formula \( F = mg \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity. This force acts downwards towards the center of mass of the Earth, and it is one of the key factors in maintaining static equilibrium in structures such as the mobile in the exercise.

Rotational Motion

Rotational motion pertains to the motion of an object around a center or a point of rotation. One of the critical aspects of rotational motion is torque, a measure of the tendency of a force to rotate an object about an axis. The torque is calculated using the formula \( \tau = rF\sin(\theta) \), where \( r \) is the distance from the pivot point to the point where the force is applied, \( F \) is the force magnitude, and \( \theta \) is the angle between the force vector and the lever arm.

For the mobile in our exercise, the rotational equilibrium is achieved when the sum of torques about the pivot point is zero. This means the torques exerted by the hanging objects and the metal bar itself must balance each other for the mobile to be level.

Static Equilibrium

Static equilibrium is a state where an object remains at rest, or its center of mass moves at a constant velocity. For an object to be in static equilibrium, two conditions must be met: the net force acting on the object must be zero (translational equilibrium), and the net torque acting on it must also be zero (rotational equilibrium).

In the context of the mobile, each hanging block and the metal bar contribute to the total force and torque. The mobile is in static equilibrium when it remains perfectly horizontal without rotating, which implies that the torques from the left and right sides around the pivot must cancel out. Calculating these torques and balancing them allows us to determine the mass of the wooden block needed on the right end to achieve static equilibrium.

Center of Mass

The center of mass is the point in an object where mass is equally distributed in all directions. It's the point where we can consider the object's entire mass to be concentrated for the analysis of translational motion. For regular, symmetric objects, the center of mass coincides with the geometric center. In irregular objects, or systems like the mobile in our exercise, its location can be determined through calculation or experimentation.

The importance of the center of mass in equilibrium problems is evident in the calculation of torque. As shown in our exercise, the metal bar’s contribution to torque was calculated using the distance from its center of mass to the pivot point. Understanding the concept of center of mass allows us to predict the behavior of the mobile and solve for the unknowns in equilibrium situations.