How to solve a problem with a bar laying against a wall

[Jay9313]

Ok, so I have completely *solved* this problem. What I mean is that I have attached the problem in a photo, and my work in a separate photo. My first answer was 62.2° and I did not round at all, I left all of the answers in my calculator. I rounded some on the second time through and got my answer down to 60.4°. My professor is adamant that the answer is 59.9° and I'm kind of a little freaked out. Could someone take a look at this and let me know if there is another way to do this? Because I had to use the law of sines and cosines and it seems like a rather challenging

 

[mfb]

Where does the third line (F_f1=F_N come from?

Why do you use 0.235 in line 4? The problem statement uses 0.25.

What does your third sketch represent and why do you have the same angle theta there?

It would be easier to follow the formulas if you did not replace µ and the sin/cos values by numerical values.

 

[Jay9313]

Where does the third line (F_f1=F_N come from?

Why do you use 0.235 in line 4? The problem statement uses 0.25.

What does your third sketch represent and why do you have the same angle theta there?

It would be easier to follow the formulas if you did not replace µ and the sin/cos values by numerical values.

I completely agree. I was really against putting numerical values in, but If I had not done that, I would have really long expressions that would clutter up my work. The .235 in line 4 is .235mg, where mg is the weight. And I did mess up the third line (F_f1=F_N Let me re-work that line really quickly and see if I get a closer answer! If this line solves my problem, I'll be really upset.

 

[Jay9313]

Without rounding, that brought my answer down to 61.8 degrees. I can try to re=work it without replacing any values with numerical values if you'd like

 

[Jay9313]

I added a new photo of my work and I tried to clean it up and show more thought progressions. I left more of the symbols in and didn't change them to numeric values, but only got so far. If anyone has a solution, please feel free to show me how to get there.

 

[Chestermiller]

I got 61.8 degrees also using 2 force balances and a moment balance. My final equation was tanθ=2 - 0.5 tan 15. I think I did it correctly. I plugged in numbers only at the very end.

Chet

 

[Jay9313]

I got 61.8 degrees also using 2 force balances and a moment balance. My final equation was tanθ=2 - 0.5 tan 15. I think I did it correctly. I plugged in numbers only at the very end.

Chet

Thank you so much! Could I trouble you to show me your work using the moment balance? I would be very grateful!

 

[Chestermiller]

Thank you so much! Could I trouble you to show me your work using the moment balance? I would be very grateful!

Let N_A be the normal force exerted by the wall at point A, and let N_B be the normal force exerted by the wall at point B. Let F be the frictional force at point B.

Horizontal Force Balance: $N_A\cos{15}=F$
Vertical Force Balance: $N_A\sin{15}+N_B=mg$
Moment Balance Around Point A: $mg\frac{L}{2}\cosθ-N_BL\cosθ+FL\sinθ=0$
where L is the length of the bar.

Solution to these equations:
$$F=\frac{\frac{mg}{2}}{\tanθ+\tan15}$$
$$N_B=\frac{mg\tanθ+\frac{mg}{2}\tan15}{\tanθ+\tan15}$$
So,
$$\frac{F}{N_B}=\frac{1}{2\tanθ+\tan15}$$

 

[Jay9313]

Let N_A be the normal force exerted by the wall at point A, and let N_B be the normal force exerted by the wall at point B. Let F be the frictional force at point B.

Horizontal Force Balance: $N_A\cos{15}=F$
Vertical Force Balance: $N_A\sin{15}+N_B=mg$
Moment Balance Around Point A: $mg\frac{L}{2}\cosθ-N_BL\cosθ+FL\sinθ=0$
where L is the length of the bar.

Solution to these equations:
$$F=\frac{\frac{mg}{2}}{\tanθ+\tan15}$$
$$N_B=\frac{mg\tanθ+\frac{mg}{2}\tan15}{\tanθ+\tan15}$$
So,
$$\frac{F}{N_B}=\frac{1}{2\tanθ+\tan15}$$

I actually considered using a moment approach such as yourself, but I didn't want to introduce a new variable, as I already had quite a few unknowns. But thank you so much, you have no idea how much I appreciate it.