• Support PF! Buy your school textbooks, materials and every day products Here!

Using Newton's Laws - A problem involving 3 masses and an incline

  • Thread starter RslM
  • Start date
  • #1
8
0
This is my first time posting. I see there is a protocol for making proper posts, so I am sorry if I make any errors.

Homework Statement



I have a problem involving a system with an incline and three masses. A picture of the system is in the attachment labeled 15.1. I am to calculate the tension in each string and the acceleration of the system.

Homework Equations





The Attempt at a Solution



I have an idea for solving this problem and would like to know if I'm on the correct track. I divided the system up into two individual systems seen in attachments 15.2 and 15.3. My reasoning is that I should be able to calculate the first string tension (T1) in system one and then calculate the second string tension (T2) in system two. I would then use the difference between T1 and T2 to derive a net force on the entire system. With a net force I could then calculate the acceleration of the system as a whole.
 

Attachments

Answers and Replies

  • #2
berkeman
Mentor
57,517
7,536
This is my first time posting. I see there is a protocol for making proper posts, so I am sorry if I make any errors.

Homework Statement



I have a problem involving a system with an incline and three masses. A picture of the system is in the attachment labeled 15.1. I am to calculate the tension in each string and the acceleration of the system.

Homework Equations





The Attempt at a Solution



I have an idea for solving this problem and would like to know if I'm on the correct track. I divided the system up into two individual systems seen in attachments 15.2 and 15.3. My reasoning is that I should be able to calculate the first string tension (T1) in system one and then calculate the second string tension (T2) in system two. I would then use the difference between T1 and T2 to derive a net force on the entire system. With a net force I could then calculate the acceleration of the system as a whole.
Welcome to the PF. Your first post is exemplary! :smile:

Your approach may well work, but I'd be inclined to draw 3 FBDs (one for each mass), and solve the resulting equations simultaneously.
 
  • #3
gneill
Mentor
20,802
2,779
What's the coefficient of friction between mass 2 and the inclined surface?

I believe that you'll have difficulty finding the individual tensions by your method, since those tensions are going to depend upon the net acceleration of the system which, in turn, depends upon all three masses.
 
  • #4
8
0
Calculations

I'm sorry, I made an error with attachment 15.3 as I forgot to add the force vector T1 to my diagram. I've corrected it in attachment 15.4. Attachment 15.5 shows how I've designated the masses and the two string tensions.

The problem states:

Three objects are connected on an inclined table as shown in the diagram (Attachment 15.1). The objects have masses 8.0 kg, 4.0 kg and 2.0 kg as shown, and the pulleys are frictionless. The tabletop is rough, with a coefficient of kinetic friction of 0.47, and makes an angle of 20.0° with the horizontal.

My calculations are as follows:

m1 = 8.0 kg, m2 = 4.0 kg, m3 = 2.0 kg, uk = 0.47, θ = 20 degrees

System 1:

m2g = Fg2 =(4.0 kg)(9.8 m/s^2) = 39.2 N
Fg2x = (39.2 N)(sin(20)) = 13.4 N
Fg2y = (39.2 N)(cos(20)) = 36.84 N

Fg2y = Fn = 36.84 N

Ffr = ukFn = (0.47)(36.84N) = 17.31 N

m1g = (8.0 kg)(9.8 m/s^2) = 78.4 N

Fnet = ma

m1g - T1 = m1a

T1 = m1g - m1a

T1 - Ffr - Fg2x = m2a

T1 = m2a + Ffr + Fg2x

m1g - m1a = m2a + Ffr + Fg2x

m1g - Ffr - Fg2x = m2a + m1a

(m1g - Ffr - Fg2x) / (m2 + m1) = a

(78.4 N - 17.31 N - 13.4 N) / (4.0 kg + 8.0 kg) = 3.97 m/s^2

T1 = m1g - m1a

m1a = (8.0 kg)(3.97 m/s^2) = 31.76 N

T1 = 78.4 N - 31.76N = 46.64 N

T1 = 46.64 N

System 2

m3g = (2.0 kg)(9.8 m/s^2) = 19.6 N

Ffr = 17.31 N (As calculated in system 1)

T1 + Ffr - Fg2x - T2 = m2a

T2 = T1 + Ffr - Fg2x - m2a

m3g - T2 = m3a

T2 = m3g - m3a

m3g - m3a = T1 + Ffr - Fg2x - m2a

(m3g - T1 - Ffr + Fg2x) / (m3 + m2) = a

(19.6 N - 46.64 N - 17.31 N + 13.4 N) / (2.0 kg + 4.0 kg) = -5.16 m/s^2

T2 = m3g - m3a

T2 = 19.6 N - (-10.32 N)

T2 = 29.92 N

Net Force : T1 - T2

46.64 N - 29.92 N = 16.72 N

Mass of entire system: m1 + m2 + m3

8.0 kg + 4.0 kg + 2.0 kg = 14 kg = M

Acceleration of system:

a = Fnet / M

a = 16.72 N / 14 kg =1.19 m/s^2

a = 1.19 m/s^2 towards the 8.0 kg mass.
 
  • #5
gneill
Mentor
20,802
2,779
Nope. Doesn't work. You're system 1 goes off the rails when you write:

T1 - Ffr - Fg2x = m2a

This equation cannot be true since it does not include the effect of T2 on the acceleration of m2. As a result your value for T1 is incorrect. That error propagates through the subsequent calculations.
 
  • #6
8
0
With this attempt, I've determined a new equation for m2 which accounts for both T1 and T2 and have incorporated all three masses into determining an acceleration for the entire system.

m2g = Fg2 =(4.0 kg)(9.8 m/s^2) = 39.2 N
Fg2x = (39.2 N)(sin(20)) = 13.4 N
Fg2y = (39.2 N)(cos(20)) = 36.84 N

Fg2y = Fn = 36.84 N

Ffr = ukFn = (0.47)(36.84N) = 17.31 N

m1g = (8.0 kg)(9.8 m/s^2) = 78.4 N

T1 -T2 - Ffr - Fg2x = m2a

m1g - T1 = m1a

T1 = m1g - m1a

T2 -m3g = m3a

T2 = m3g + m3a

(m1g - m1a) - (m3g + m3a) - Ffr - Fg2x = m2a

m1g - m1a - m3g - m3a - Ffr - Fg2x = m2a

((m1 - m3)g - Ffr - Fg2x) / (m1 + m2 + m3) = a

(6 kg (9.8 m/s^2)) - 17.296 N - 13.41 N / 14 kg = 2.0 m/s^2

Acceleration of the system: 2.0 m/s^2 towards the 8.0 kg mass.
 
  • #7
gneill
Mentor
20,802
2,779
Your acceleration value looks good :approve:
 
  • #8
8
0
Awesome! I'm very glad.

Thank you both for the assistance and for the welcome to this forum. Much obliged. :)
 
  • #9
3
0
Im confused with the calculation for T1:

"m1g - T1 = m1a

T1 = m1g - m1a"

why in this case is the tension subtracted from the gravity force and not vice versa? Thanks!
 
  • #10
3
0
also, how do you calculate which direction the system is moving?
 

Related Threads on Using Newton's Laws - A problem involving 3 masses and an incline

Replies
5
Views
1K
Replies
7
Views
2K
Replies
4
Views
725
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
11
Views
6K
  • Last Post
Replies
8
Views
495
  • Last Post
Replies
4
Views
2K
Replies
5
Views
3K
  • Last Post
Replies
3
Views
2K
Top