Why Do Reactions Differ in Problem P16.3 Compared to Standard Statics?

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SUMMARY

The discussion centers on the differing approaches to solving Problem P16.3 in statics, specifically regarding the reactions at points A and B on an inclined plate. The original poster asserts that reactions should align with the incline's angle, theta, while others clarify that the author uses horizontal and vertical components for the reactions. At point A, the reaction force includes both horizontal (friction, F) and vertical (normal force, N_A) components, whereas at point B, the reaction is purely horizontal (normal force, N_B) due to the smooth surface.

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cipotilla
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Hello, there something that doesn't make sense to me in the attached problem, P16.3. You can see from the diagram that you will have two reculting reactions, RA and RB. According to what I know from statics, the reactions will be at the same angle theta as the incline of the plate, see my solution on page 2. Why does the book's author, solve the problem using reactions that are purely horizontal and purely vertical?

Thanks.
 

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cipotilla said:
According to what I know from statics, the reactions will be at the same angle theta as the incline of the plate, see my solution on page 2.
Why do you think that?
Why does the book's author, solve the problem using reactions that are purely horizontal and purely vertical?
He doesn't. At the A end of the plate, the reaction force of the surface on the plate has both horizontal (friction, F) and vertical (normal force, N_A) components. The author identifies the horizontal component as friction, but it's still part of the overall force that the surface exerts on the plate.

At the B end, the surface is smooth so there is no vertical component of reaction force, only the horizontal normal force (N_B).
 

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