Determining Normal Forces for Particle P

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SUMMARY

The discussion focuses on calculating the normal forces acting on particle P, which has a mass of 1 kg and experiences an upward acceleration of 2.46 m/s². The normal force exerted by the slot (Nslot) is determined to be 9.8 N, corresponding to the weight of the particle. The normal force from the rod (Nrod) is calculated using the equation Nrod cos 30° = m*a, resulting in Nrod = 2.84 N. The participants clarify the directions of the forces and the importance of considering the weight of the particle in the calculations.

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  • Understanding of Newton's second law (F = ma)
  • Knowledge of normal force concepts in physics
  • Familiarity with vector projections and trigonometric functions
  • Basic principles of dynamics involving rotating systems
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songoku
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Homework Statement


Particle P has a mass of 1 kg and is confined to move along the smooth vertical slot due to the rotation of the arm AB. Assume that at the instant as shown, the acceleration of particle P is 2.46 ms-2 upward. Determine the normal force Nrod on the particle by the rod and the normal force Nslot on the particle by the slot in the position shown. Friction forces are negligible.

slot.jpg



Homework Equations


F = ma



The Attempt at a Solution


I think that the Nslot is directed out of the page and equals to W so Nslot = 1*9.8 = 9.8 N

Then, Nrod is directed to the top left and is perpendicular to AB. So, Nrod can be projected to Nrod cos 30o directed vertically. Hence :
Nrod cos 30o = m*a
Nrod = 2.84 N

But there is no such answer in the multiple choices...

Thanks
 
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Nslot is normal to the walls of the slot, so it is in-plane and horizontal. As for Nrod, you are right, Nrod is perpendicular to AB, but there is the weight of the body which is vertical and points downward.

ehild
 
Hi ehild

Ok I think that Nslot is the projection of the normal force in horizontal direction and yes I forgot to take the weight into account. I think I figure out the answer

Thanks a lot !
 

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