Question

A uniform beam of length L = 3 meters and mass M = 150 kg has a block of mass m = 300 kg and negligible size firmly attached to one end as shown below. The beam is attached by a hinge to the wall and extends perpendicularly from the wall. It is braced with a rope attached some distance away, which makes an angle θ = 30° with respect to the beam.


What is the tension T in the rope?


The rope is cut. What is the magnitude |α| of the angular acceleration of the beam and block immediately after the rope is cut?

Answer 1

A diagram would be very helpful.

Answer 2

This is a tricky problem and calls for calculus! By the definition of torque, we have \( \tau = I \alpha\) for net torque \(\tau\). When the beam is being held by the rope, we obviously have \(\alpha = 0\) since the beam is not accelerating (in fact, it's not moving!). Thus, we know the net torque on the beam is zero, and thus the magnitudes of the torques applied by both the rope and gravity on the beam are equal. Recall also that torque can be expressed in terms of force by \(\vec \tau = \vec r \times \vec F\). Therefore, \(\tau = rF\sin \theta\) for angle \(\theta\) between \(\vec r\) and \(\vec F\). Since the force by the rope is \(T\) Here, \(r\) is the length of the board \(L\) since the rope is attached at the opposite end of the board that is attached to the wall, we can simply substitute as follows.\[\tau_{\text{rope}} = LT\sin \theta\]Now, the torque applied by gravity on the board is more difficult to calculate. Let us split this into two independent torques. First we will consider the torque by gravity on the board \(\tau_{\text{board}}\) and the torque by gravity on the mass at the end of the board \(\tau_{\text{mass}}\). When \(\vec r\) and \(\vec F\) are perpendicular, which we have in the gravity case since the board is horizontal while gravity points vertically, we have \(\tau = rF\). For the mass at the end of the board, the gravitational force is simply \(F = mg\). Hence, the torque by the mass is \(\tau_{\text{mass}} = Lmg\). Since mass is distributed along the board at different distances from the axis of rotation, we must perform an integral along the entire board to figure out the net torque on the board due to gravity. Let \(dm\) be an infinitesimal mass element of the board, and \(d\ell\) be an infinitesimal length element of the board. Now, since the mass density is constant throughout the boar, we can express \(\frac{dm}{d\ell} = \frac{M}{L}\), or \(dm = \frac{M}{L}d\ell\). The force by gravity on this infinitesimal mass element is \(dF=gdm = \frac{Mg}{L} d \ell\), so the infinitesimal torque on this mass element a distance \(\ell\) from the axis of rotation is \(d\tau_{\text{board}} = \ell dF = \frac{Mg}{L}\ell d\ell \). Integrating both sides along the entire length of the board gives us the following.\[\tau_{\text{board}} = \frac{Mg}{L} \int_0^L \ell d \ell = \frac{Mg}{L} \cdot \frac{L^2}{2} = \frac{1}{2}MgL \] Thus, we can write the following expression for the magnitude of the torque applied by gravity on the system \(\tau_{\text{gravity}}\). \[\tau_{\text{gravity}} = \tau_{\text{mass}} + \tau_{\text{board}} = Lmg + \frac{1}{2}MgL = Lg\left(m + \frac{M}{2}\right)\]Setting \(\tau_{\text{gravity}} = \tau_{\text{rope}}\) allows us to write the following expression for the tension in the rope \(T\). \[\boxed{\displaystyle T = \frac{g}{\sin \theta}\left(m + \frac{M}{2}\right)}\]When the rope is cut, we can disregard \(\tau_{\text{rope}}\) and only consider \(\tau_{\text{gravity}}\). Since \(\tau = I \alpha\), we have \(\alpha = \frac{\tau_{\text{gravity}}}{I}\). The moment of inertia of the beam is found by adding the moments of inertia of the mass and the board \(I = I_{\text{board}} + I_{\text{mass}}\). For a point mass, we have \(I = MR^2\). Thus, for the mass at the end of the board, we have \(I_{\text{board}} = mL^2\). For the board, we will integrate once again, considering an infinitesimal mass element \(dm\) a distance \(\ell\) from the axis of rotation. Thus, \(dI_{\text{board}} = \ell^2 dm = \frac{M}{L} \ell^2 d\ell\). Performing an integral across the whole length of the board gives the following.\[I_{\text{board}} = \frac{M}{L} \int_0^L \ell^2 d\ell = \frac{M}{L} \cdot \frac{L^3}{3} = \frac{ML^2}{3}\]Thus, we have the moment of inertia of the system. \[I = I_{\text{board}} + I_{\text{mass}} = \frac{ML^2}{3} + mL^2 = L^2 \left(m + \frac{M}{3}\right)\]Therefore, the angular acceleration of the board is given as follows.\[\alpha = \frac{\tau_{\text{gravity}}}{I} = \frac{Lg\left(m + \frac{M}{2}\right)}{L^2 \left(m + \frac{M}{3}\right)} = \boxed{\displaystyle \frac{g}{L} \cdot \frac{m + \frac{M}{2}}{m + \frac{M}{3}}}\]

Answer 3

Thank you so much!!! I think the formulas are slightly off but thanks for explaining so well! it was very helpful!

Answer 4

Yakeyglee, thanks for the detailed answer. I came to this question confused, and your answer helped quite a bit.

I am still confused by one aspect though: shouldn't the force of gravity be a function of theta? E.g., if the board were parallel to the direction of gravity (either pointing directly up or hanging straight down on the hinge), then the force would be zero, correct? It seems like the force should be something like: \[g \times sine(\Theta)\]
I am just thinking aloud here.