Problem on an inclined plane

  • Thread starter haengbon
  • Start date
  • #1
38
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Homework Statement



A 3kg load on an inclined plane of angel 30 degrees is connected by a cord over a frictionless pulley to a second load which has a mass of 2 kg. Find the:

a) Acceleration of each load
b) tension of each chord


Homework Equations



none?


The Attempt at a Solution



block A / box on inclined plane

[tex]\sum[/tex]Fy= 0
-Wycos30[tex]\circ[/tex] + NF= 0
-30cos30 + NF = 0
-25.98 + NF = 0
-25.98 = NF
NF = 25.98

f=[tex]\mu[/tex] NF
f= 0.3 (25.98)
f= 7.794

[tex]\sum[/tex]Fx= ma
Wxsin30[tex]\circ[/tex] - T - f = ma
15 - T - 7.794 = ma
15 - 7.794 - T = (3)(a)
-T = 3a - 7.206
T = 7.206 - 3a

block b

[tex]\sum[/tex]Fx= ma
T-W = ma
T - (2)(10) = (2)a
T - 20 = 2a
T= 2a+20

T=T
2a+20 = 7.206 - 3a
2a+3a = 7.206 - 20
5a = -12.794
a = - 2.56 m/s2

T1=7.206 - 3a
T= 7.206- 3(-2.56)
T= 14.886 N

T2= 2a+20
T= 2(-2.56) + 20
T= -3.1176 N

T2+T1 = TotalT
14.886 + (-3.1176) = TotalT
TotalT= 11.7684 N

I'm not sure if I got the right answer :( someone help?
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
7
Loads are connected by a single cord. So T1 = T2 = T.
 
  • #3
38
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Loads are connected by a single cord. So T1 = T2 = T.
so the tensions I got are incorrect then? :frown:
may I ask if the acceleration is correct?
 
  • #4
rl.bhat
Homework Helper
4,433
7
Your acceleration is correct.
 
  • #5
38
0
Your acceleration is correct.
thank you ! :) so I only need to use one equation for my tension, and that's the answer? thank you very much again :) I really appreciate it ^^
 

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