Friday

April 18, 2014

April 18, 2014

Posted by **Amy** on Monday, May 21, 2012 at 5:05pm.

(b) The interaction partner of this normal force has what magnitude and direction?

Magnitude 87.46 N

Direction 61° below the horizontal?

c) What is the static frictional force exerted on the crate by the ramp?

48.48 N

(e) The normal and frictional forces are perpendicular components of the contact force exerted on the crate by the ramp. What is the magnitude and direction of the contact force?

Magnitude 99.99 N

Direction ° above the horizontal?

All I need to know is the answer and explanation for e) direction in degree because I got tan-1 (87.46/48.48) I got 60.99 and 60.99-29= 31.99 degree but its wrong I tried 60.99 and 31.99 both are wrong please someone help me out. Thank you!

- Physics -
**Elena**, Monday, May 21, 2012 at 5:30pmContact force = sqrt(87.46² + 48.48²) = sqrt (7679+2350) = 9999.6 = 10000,

tan φ = N/F(fr) = 87.46/48.48 = 1.8,

φ = 60.99=61º.

Direction is 61º +29º=90º.

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