Two children hang by their hands from the same tree branch. The branch is straight, and grows out from the tree trunk at an angle of 27.0° above the horizontal. One child, with a mass of 40.0 kg, is hanging 1.80 m along the branch from the tree trunk. The other child, with a mass of 29.0 kg, is hanging 2.00 m from the tree trunk. What is the magnitude of the net torque exerted on the branch by the children? Assume that the axis is located where the branch joins the tree trunk and is perpendicular to the plane formed by the branch and the trunk.
Mathematics · High School · Thu Feb 04 2021
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To calculate the magnitude of the net torque exerted on the branch by the children, we can use the formula for torque (\( \tau \)) about a pivot point, which is given by:
\[ \tau = r \times F \times \sin(\theta) \]
where \( r \) is the distance from the pivot point to the point where the force is applied (which, in this case, is the child's distance from the trunk), \( F \) is the force applied (which is the weight of the child), and \( \theta \) is the angle between the force vector and the lever arm (the branch in this case).
The force due to gravity (weight) acting on each child can be calculated using the formula
\[ F = m \times g \]
where \( m \) is the mass of the child and \( g \) is the acceleration due to gravity (approximately \( 9.81 m/s^2 \)).
For the first child (40.0 kg) hanging 1.80 m from the trunk:
\[ F_1 = m_1 \times g \] \[ F_1 = 40.0 \, kg \times 9.81 \, m/s^2 \] \[ F_1 = 392.4 \, N \]
The torque exerted by this child is:
\[ \tau_1 = r_1 \times F_1 \times \sin(\theta) \] \[ \tau_1 = 1.80 \, m \times 392.4 \, N \times \sin(27.0^\circ) \] \[ \tau_1 = 1.80 \, m \times 392.4 \, N \times 0.454 \] (since \(\sin(27.0^\circ) \approx 0.454\)) \[ \tau_1 = 319.16 \, Nm \]
For the second child (29.0 kg) hanging 2.00 m from the trunk:
\[ F_2 = m_2 \times g \] \[ F_2 = 29.0 \, kg \times 9.81 \, m/s^2 \] \[ F_2 = 284.49 \, N \]
The torque exerted by this child is:
\[ \tau_2 = r_2 \times F_2 \times \sin(\theta) \] \[ \tau_2 = 2.00 \, m \times 284.49 \, N \times \sin(27.0^\circ) \] \[ \tau_2 = 2.00 \, m \times 284.49 \, N \times 0.454 \] \[ \tau_2 = 258.91 \, Nm \]
The total torque is the sum of the torques due to each child:
\[ \tau_{total} = \tau_1 + \tau_2 \] \[ \tau_{total} = 319.16 \, Nm + 258.91 \, Nm \] \[ \tau_{total} = 578.07 \, Nm \]
Therefore, the magnitude of the net torque exerted on the branch by the children is \( 578.07 \, Nm \).
Extra: Torque is a measure of the turning effect of a force applied to a rotational system at a distance from the axis of rotation. In physics, it is essential when dealing with rotational dynamics. The branch's angle in this problem creates a situation where not all of the gravitational force contributes to the torque; only the component perpendicular to the branch does. That's why we multiply by the sine of the angle - it gives us the right component of the force that contributes to the rotation. Torque problems often involve determining the sum of all torques to understand how a system will rotate, and whether it will stay in equilibrium or not. Mechanisms in equilibrium have a net torque of zero.