coursehero_TEST 2 STUDY GUIDE ch5

# coursehero_TEST 2 STUDY GUIDE ch5 - PHYSICS TEST 2 STUDY...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PHYSICS TEST 2 STUDY GUIDE: CHAPTER 5: NEWTON’S LAWS (Equilibrium, Free Body Diagram) Q1: The force up will be T 2 and the force left will beT 1 . For the diagonal tension, you want to break it up into its x and ycomponents. The force right will be 100cos30 and the force down will be100sin30 and mg. We are going to say mg = 0 because it is a very light ring. Since the ring is stationary, acceleration will be 0. F x = ma = 0 = 100cos30 - T 1 (it doesnot matter which one you make negative since net force is 0) F y = ma = 0 = 100sin30 - T 2 So T 1 = 100cos30 And T 2 = 100sin30 100*cos30= T 1= 86.6 N 100*sin30= T2=50 N Q2: The forces in the figure are acting on a 7.8 object. Find the value of , the x-component of the object's acceleration. Find the value of , the y-component of the object's acceleration. F = ma F x = ma x WHERE: F x = forces in the x direction m = mass a x = acceleration in the x direction If you look at the figure, there are only two forces in the x direction. The first force is the x component force 4.0 N acting If you look at the figure, there are only two forces in the x direction....
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

coursehero_TEST 2 STUDY GUIDE ch5 - PHYSICS TEST 2 STUDY...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online