Determine force exerted on rod & reaction

[JJBladester]

Homework Statement

A conveyor system is fitted with vertical panels, and a 300-mm rod AB of mass 2.5kg is lodged between two panels as shown. Knowing that the acceleration of the system is 1.5m/s² to the left, determine (a) the force exerted on the rod at C, (b) the reaction at B.

Answers:
(a)3.43N at 20 degrees above the horizontal (1st quadrant)
(b)24.4N at 73.4 degrees above the horizontal (2nd quadrant)

Homework Equations

[tex]\sum F_x=m\bar{a}_x[/tex]

[tex]\sum F_y=m\bar{a}_y[/tex]

[tex]\sum M_G=\bar{I}\alpha[/tex]

Where [itex]G[/itex] is the center of mass and [itex]\bar{x}[/itex] and [itex]\bar{y}[/itex] are the x and y coordinates of G.

The Attempt at a Solution

Draw a free body diagram:

Sum of the forces equations:

[tex]\sum F_x=Ncos(20)-Bcos(70)=ma[/tex]

[tex]\sum F_y=Nsin(20)-mg+Bcos(70)=0[/tex]

That's all I can think of and I know that assuming B's reaction is at 70 degrees is wrong because the answer states that it is at 73.4 degrees above the horizontal (2nd quadrant).

What is wrong with my FBD?

 

[anathema]

Thank you for sharing your thoughts and attempt at solving this problem. Your free body diagram and equations seem to be correct. However, there may be a minor mistake in your calculation for the reaction at B. Instead of using the angle of 70 degrees, you can try using the angle of 20 degrees, which is the complementary angle to 70 degrees. This should give you the correct answer of 73.4 degrees for the reaction at B. Keep up the good work!