MHB Solve Q5 Part II Maths Exam: Normal Reaction Force on C

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The discussion focuses on resolving the normal reaction force on point C in a physics exam question. The user calculated the tension in the string as 2.24N but arrived at a normal reaction force of 5.96N, which differs from the provided solution of 5.26N. A key point raised is the potential neglect of the frictional force at C, which affects the calculations. The correct approach involves resolving forces perpendicular to the plane, where the frictional force does not contribute. Clarification on the question number was also made, confirming it was indeed question 6.
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So I have my final maths exam tomorrow and thought I'd do a few new past papers to brush up.

Here is my problem:
Q 5) part ii....

I worked out the tension in the string which was correct at 2.24N, but when I resolved vertically to find out the normal reaction force on C, I came out with 5.96N. In the solutions they resolved perpendicular to the plane and ended up with 5.26N.
My workings were: T+NCos(20)=0.8g

Any ideas on where I went wrong?

Thanks!
 

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I have moved this topic, as it is more physics than geometry.

Just to verify, you are actually referring to question 6, correct?
 
Yes! Sorry Q6) part ii (Blush)
 
It looks as though you may be neglecting the frictional force at C, which acts in the direction CB (so as to balance the horizontal component of the normal force at C). If you resolve perpendicular to the rod then the frictional force has no component in that direction.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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