(II) An object is hanging by a string from your rearview mirror. While you are decelerating at a constant rate from 25 m/s to rest in 6.0 s, ($a$) what angle does the string make with the vertical, and ($b$) is it toward the windshield or away from it? [$Hint$: See Example 4-15.]

Answer

a)$23^{\circ}$
b)toward

October 27, 2020

Finally, the answer I needed, thanks Averell H.

Top Physics 101 Mechanics Educators

Christina K.

Rutgers, The State University of New Jersey

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Video Transcript

An object is hanging by a string from your rear view mirror. All you're decelerating at a constant rate of 25m/s to rest in six seconds. What angle does the string make with the vertical? Is it towards the windshield or away from it? Okay, so just to get us a free body diagram here is the object. It has a weight that will pull it straight down four. Stupid tension will hold it up and then let's let this angle be said of. So let's say that since the car is moving to the right and is slowing down, the acceleration and the net forces are going to be pointed to the left. So the acceleration in the X direction is v minus V, not over teeth, Which is 0 -25 meters per second over six seconds, Which is negative 4.17 leaders per second squared. So Using Newton's 2nd Law, we know that the force of the tension in the X. Direction. Sorry, the Y direction is F. T. Co sign of Ceta. And it needs to equal to the weight to be an equilibrium. So if we're looking in the X direction negative, the four suit attention sign of data is equal to the mass times the acceleration in the X. Direction. So we have that mass times acceleration. The X direction is negative FT. Sine of theta which is negative MG over co sign of data. So solving for FT in this equation minus theta, just negative MG. Tangents of data. So the masses will cancel the acceleration and the X direction is negative G. Attention to feta. We just solved for A. X. We know what G. Is. So solving for the angle And we get 23° so it is going towards the windshield.

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