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Angle of pendulum in accelerated incline

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Consider the following system consisting of a box sliding down a plane. The coefficient of friction between the plane and the box is $\mu$. A pendulum is attached to the top of the box as shown.

The acceleration of the box+pendulum is $a=g(\sin\theta-\mu\cos\theta)$ I believe.
In a non-inertial frame attached to the box, the free-body diagram for the pendulum is

My goal is to find the angle $\phi$ from the equilibrium of these 3 forces. I have to pick x and y axes to decompose these forces. If I pick the x axis along the fictitious force and the y perpendicular to it I get (I think)

$$m_P a+T\sin\phi=m_P g\sin\theta$$

and

$$T\cos\phi=m_{P}g\cos\theta$$

which I can then solve for $\phi$ :

$$\tan\phi=\frac{g\sin\theta-a}{g\cos\theta}=\mu$$

Is this right?

asked Sep 28 in Physics Problems by member192 (270 points)
edited Sep 28 by member192
Yes your result is correct. It is surprising that it does not involve the angle $\theta$. There ought to be an intuitive explanation for this, but at present I cannot think what it is.
I fixed the sign ($-\mu$ to $\mu$)
This answer must be suspect as it predicts an inclination of the pendulum when the object is on the flat with $\theta=0$. I think that the error is in the first line where the acceleration of the box+pendulum has been found where the normal reaction force used to evaluate the frictional force is incorrect.  That normal reaction depends upon the tension in the string and hence the angle $\phi$ and well as $\theta$ and the weight of the bob.
@Farcher I do not follow you. Please could you elaborate? Perhaps post the correct solution?

I think the equation $a=g(\sin\theta-\mu\cos\theta)$ assumes that $\tan\theta \ge \mu$. Otherwise $a$ would be negative when we know that the block is not moving. So the result $\tan\phi=\mu$ only applies for $\tan\theta \ge \mu$.
I take your point about the minimum possible value for the slope of the incline. The frictional force is controlled by the normal reaction on the box.  The force that the box and the pendulum exerts on the slope is not $(m_b+m_p) g \cos \theta$ because the box has no knowledge of the force $m_pg$ . All the  box "feels" is the tension in the string which is not vertical and hence relating the calculated component of the tension in the string which is perpendicular to the slope to just $\theta$ is incorrect.
@Farcher Surely if the pendulum is not accelerating perpendicular to the slope then the normal reaction between the box and slope is $(m_b+m_p)g\cos\theta$. The slope does not know (and does not care) whether the pendulum is attached to the box by a string or glued directly to it. All it "feels" is the perpendicular component of the total weight resting on it. It does "feel" the force $m_p g\cos\theta$ via the tension $T$ which is just an internal force for the box-pendulum system.

Just as a check I treated box and pendulum as separate Free Bodies linked by tension $T$in the string. As expected, I got the same acceleration $a=g(\sin\theta-\mu\cos\theta)$.

If you still think this is wrong please would you post your calculation of what $a$ and $\tan\phi$ should be?

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