Data 2) ニ 80N %3D 0=30° my = 3kg F= 80N 0= 30° Calculate: a) acceleiation ) Tension. Me= 0.10 FN Fx=Feos30° Fx =(80N)Cos: =F sin 30° /- =69.3 N Solation Fy =(30N) Sin3 = 40N2 Bexng Box mi Box m2 | ZFx=max = Fx=FT EFx = Fr,= m2 az 69:3N - 17.3 im/s mi aky * F L F because G acts to accelera ケニm2a (Box m) only * Fr=51.9N LE
Data 2) ニ 80N %3D 0=30° my = 3kg F= 80N 0= 30° Calculate: a) acceleiation ) Tension. Me= 0.10 FN Fx=Feos30° Fx =(80N)Cos: =F sin 30° /- =69.3 N Solation Fy =(30N) Sin3 = 40N2 Bexng Box mi Box m2 | ZFx=max = Fx=FT EFx = Fr,= m2 az 69:3N - 17.3 im/s mi aky * F L F because G acts to accelera ケニm2a (Box m) only * Fr=51.9N LE
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Please solve this exercise. I don't think I did it right.

Transcribed Image Text:Data
2-
4kg
my =3Kq
F= 80 N
8=30°
0=30°
Calculate:
Me=0.10
a-) acceleiation
6) Tension.
30°
Fx=Fcos30° Fy=(80N)CoS 30
ty=F sin 30°
=69.3 N
Solution
mgRng
vmg
Fy =(30N) Sin30
4ON
Borng
2.
Box mi
Box m2 a.
ZFy=max
Zfy = Fx-FT=ma
IFx = Fr,= m2 az
69.3N"
: 17.3 im/s
mi
* F LF because acts to accelerate
(Box m2) only
FT
ause acts To accelerate
2.
FT=51.9N
年。
Expert Solution
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