Required information Problem 16.048 - DEPENDENT MULTI-PART PROBLEM - ASSIGN ALL PARTS NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. A uniform slender rod AB rests on a frictionless horizontal surface, and a force P of magnitude 0.25 lb is applied at A in a direction perpendicular to the rod. Assume that the rod weighs 2.2 lb. Problem 16.048.a - Acceleration A of slender rod on frictionless surface Determine the acceleration of Point A. (You must provide an answer before moving on to the next part.) The acceleration of Point A is 14.6 14.6 Correct ft/s2 →.

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kollybaba55 kollybaba55 · Answer 1 · 2020-04-23T12:25:30+02:00

NOTE: The diagram is attached to this solution

Answer:

The acceleration of point A = 14.64 ft/s²

Explanation:

By proper analysis of the diagram, acceleration of point A will be: (Check the free body diagram attached)

[tex]a_{A} = \bar{a} + \frac{\alpha L}{2}[/tex]

Weight, W = mg

g = 32.2 ft/s²

m = W/g

[tex]p = m \bar{a}\\p = w \bar{a} /g[/tex]

[tex]\bar{a} = pg/w\\\bar{a} = 0.25g/2.2\\\bar{a} =3.66[/tex]

[tex]\frac{pL}{2} = I \alpha[/tex]

but [tex]I = \frac{wL^{2} }{12g}[/tex]

[tex]\frac{pL}{2} = \frac{\alpha wL^{2} }{12g}\\\alpha = \frac{6 g p}{wL}\\\alpha = \frac{6*g*0.25}{2.2L} \\\alpha = 21.96/L[/tex]

[tex]a_{A} = 3.66 + \frac{(21.96/L ) * L}{2}\\a_{A} = 3.66 + 10.98\\a_{A} = 14. 64 ft/s^{2}[/tex]