Scribd
Upload
56 views5 pages

CORREC

This document contains calculations to determine various forces, accelerations, and tensions in a system involving two masses on an inclined plane. The key results are: 1) The acceleration of the two masses down the plane is calculated to be 2.3556 m/s^2. 2) Integrating this acceleration gives the velocity of the masses as 6.8637 m/s after 2.9138 seconds. 3) Several force equations are set up and solved, determining the normal force on the first mass to be 588.4198 N and the force causing movement down the plane to be 890.4151 N. 4) Tensions in ropes connecting the two masses are calculated to

Uploaded by

Ronald Catagña
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
56 views5 pages

CORREC

This document contains calculations to determine various forces, accelerations, and tensions in a system involving two masses on an inclined plane. The key results are: 1) The acceleration of the two masses down the plane is calculated to be 2.3556 m/s^2. 2) Integrating this acceleration gives the velocity of the masses as 6.8637 m/s after 2.9138 seconds. 3) Several force equations are set up and solved, determining the normal force on the first mass to be 588.4198 N and the force causing movement down the plane to be 890.4151 N. 4) Tensions in ropes connecting the two masses are calculated to

Uploaded by

Ronald Catagña
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
You are on page 1/ 5

∑ Fy=0

N 2−w 2 y =0
N 2=w 2 y
N 2=m 2∗g∗cos( 30° )

∑ F x=m2∗a 1
w 2 x−fr 2=m2∗a1
m 2∗g∗sen (30 °)−μ 2∗(m 2∗g∗cos ⁡(30 °))=m 2∗a 1
g∗sen (30 °)−μ 2∗(g∗cos ⁡(30 °))=a 1

9,807 m/ s2∗sen( 30° )−0,3∗( 9,807 m/s 2∗cos ⁡(30 °))=a 1

a 1=2,3556 m/ s2

dv
=a 1=2,3556 m/s 2
dt
dv =2,3556 dt
v t

∫ dv=2,3556∫ dt
0 0

v=2,3556∗t

dr
=v =2,3556∗t
dv
dr =2,3556∗t dt
r t

∫ dr=2,3556∗∫ t dt
0 0

2,3556∗1 2
r= t
2
r =10
2,3556∗1 2
10= t
2
t=2,9138 s
Reemplaza el tem en la ecu v

v=2,3556∗(2,9138 s)
v=6,8637 m/s

∑ Fy=0
N 2−F 2 y−w 2 y =0
N 2=F 2 y +w 2 y
N 2=m 2∗a∗sen(30 ° )+ m2∗g∗cos ⁡(30 ° )
N 2=5 a+ 84,9311(1)

N 2=5(10,4066 m/ s2 )+ 84,9311(1)
N 2=136,9641 N

∑ F x=0
F 2 x−fr 2−w 2 x=0
m 2∗a∗cos ( 30 ° )−μ 2∗N 2−m2∗g∗sen( 30° )=0
10∗a∗cos ( 30 ° )−0,3∗(5 a+ 84,9311)−10∗(9 , 8 07)∗sen (30° )=0
8,6603 a−1,5 a−25,4793−49,035∗sen ( 30° ) =0

a=10,4066 m/s 2

∑ Fy=0
N 1+ fr 2−w 1−N 2 y=0
N 1=−fr 2+w 1+ N 2 y
N 1=−μ 2∗N 2∗sen ( 30 ° ) +m1∗g+ N 2∗cos ⁡(30 ° )
N 1=−(0,3)∗(136.9641)∗sen ( 30 ° ) +(50)∗(9.807)+( 136.9641)∗cos ⁡(30 °)
N 1=588,4198 N

∑ F x=(m 1+ m2)∗a
F−fr 1−fr 2 x−N 2 x=(m1+m 2)∗a
F=0,3∗588,4198+0,3∗136.9641∗sen (30 ° ) +136.9641∗cos ⁡( 60° )+(10+50)∗10,4066
F=890,4151 N
∑ Fy=0
T 1 y−T 2 y−w=0
T 1∗sen (23,5782°)−T 2∗sen ( 23,5782° )−m∗g=0
T 1∗sen ( 23,5782 ° )−T 2∗sen (23,5782 ° )−2∗9,807=0
T 1=44,9414+T 2

∑ Fc=0
Fc−T 1 x−T 2 x =0

m∗w2∗R−T 1∗cos ⁡(23,5782° )−T 2∗cos ⁡(23,5782° )=0

2∗( 9,4248 )2∗(9,1652)−( 44,9414+T 2)∗cos ⁡( 23,5782°)−T 2∗cos ⁡(23,5782°)=0


T 2=863,7700 N
T 1=912,8049 N
T1==10,99N

a=101,9114 m/s2
T1=8,84561
T1=8,84561N

A2=101,9114 m/s2

You might also like