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

[Needhelp2]

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!

 

[MarkFL]

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

Just to verify, you are actually referring to question 6, correct?

 

[Needhelp2]

Yes! Sorry Q6) part ii (Blush)

 

[Opalg]

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.

 

[CellCoree]

Hi there,

First of all, congratulations on taking the initiative to practice past papers before your final exam! That is a great way to prepare and ensure you are confident in your understanding of the material.

In regards to your problem, it seems like you have made a small error in your calculation. When resolving vertically, you should have used the sine function instead of the cosine function. This is because the normal reaction force is perpendicular to the inclined plane, not parallel to it. So your equation should have been: T + Nsin(20) = 0.8g.

Using this equation, you should get N = 5.26N, which matches the solution provided.

Remember to always carefully consider the direction and orientation of forces when resolving them. Keep up the good work and good luck on your exam tomorrow!