2009: Problem 24

Topic: Dynamics
Concepts: Inclined plane, Statics, Torque from weight


Solution:

Recall the slip angle on an inclined plane is given by

tanθslip=μs\tan\theta_{\text{slip}}=\mu_s

To find the tip angle, note that when the block is about to tip over, the only point in contact with the plane is the bottom corner. If we choose this as our pivot point, then the block’s weight is the only force that contributes to the torque.

For the block to stay at rest, there should be no torque. This means the weight vector (acting at the center of mass) should pass through the pivot point to have zero moment arm. By using similar triangles in the figure, we find

tanθtip=b/2a/2=ba\tan\theta_{\text{tip}}=\frac{b/2}{a/2}=\frac{b}{a}

To tip over first, we need

θtip<θslip\theta_{\text{tip}}<\theta_{\text{slip}}
tanθtip<tanθslip\tan\theta_{\text{tip}}<\tan\theta_{\text{slip}}
ba<μs\frac{b}{a}<\mu_s

so the answer is C.

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