• Support PF! Buy your school textbooks, materials and every day products via PF Here!

It's that time again: Help by tonight

Thanks in advance for the help. I'm stuck on problem #12(please see the attached file). I did this:
r= cos(26)*2.8(is this right?)
r= 2.52
a= mg/(cos(ang))
r*a=v(squared)

a= 5.9(9.8)/cos(26)
a= 64.3

v= sqrt(2.52*64.3)
v= 12.7

What's wrong with this?
 

Attachments

HallsofIvy

Science Advisor
Homework Helper
41,712
876
"r= cos(26)*2.8(is this right?)"

No, it's not. r is the "opposite" side of the right triangle:
r= sin(26)*2.8
 

HallsofIvy

Science Advisor
Homework Helper
41,712
876
"a= mg/(cos(ang))"

No, that's wrong. For one thing, there is no "m" in the acceleration. For another thing, the acceleration you want, in order to calculate speed around the circle is the acceleration toward the center, that's directed along the opposite side of your triangle, not the hypotenuse. The formula you want is a/g= tan(ang) (a= opposite side, g= near side).

You do want v2= r* a which is
v2= (2.8 sin(26))(9.8 tan(26)).

I get v= 2.42 m/s.
 

Related Threads for: It's that time again: Help by tonight

  • Posted
Replies
1
Views
1K
  • Posted
Replies
15
Views
2K
Replies
1
Views
2K
Replies
1
Views
5K
Replies
12
Views
2K
Replies
11
Views
2K
Replies
7
Views
1K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top