How to find tension in strings pulling on carts.

Click For Summary
To find the tension in strings connecting carts, start by determining the acceleration of the system using the net force and total mass. The first cart experiences a net force of 30 N minus the tension, leading to the equation 30 - T = 5 kg * 2 m/s². For the second cart, apply the same principle, considering the tension from the first string and the force acting on it. The third cart follows similarly, factoring in the tension from the second string. Drawing free body diagrams for each cart is crucial for visualizing forces and solving for tension accurately.
drunkenrobot
Messages
4
Reaction score
0

Homework Statement



3 carts are connected by strings. The first cart is 5kg, the second 2kg, and the third 8kg. A string is fixed at the front of the first cart is pulled with a force of 30 N. In the following problems be sure to illustrate your answers with the corresponding free body diagrams.

a. What is the acceleration of the train of carts?

b. What is the tension in the string connecting the first and second car?

c. What is the tension in the string connecting the second and third car?


Homework Equations



Fnet = ma

I don't remember the equations to find tension

The Attempt at a Solution



I figured out the acceleration easily:

Fnet = (m1 + m2 + m3)a
30 N = (15kg)a
a = 2 m/s ^2

The problem I'm having is I forget how to find the tension force in the strings.

Any help?
 
Physics news on Phys.org
Drawing a FBD for each cart is THE FIRST STEP and is VERY IMPORTANT.
 
ok, so for the first one I have 30 N pulling it in one direction (let's say to the left). So then the tension Force (Ft) is pulling in the opposite direction, right? But this is where I get confused. I don't know how to find the magnitude of the tension force.
 
Try to use Fnet = ma for the FIRST cart.
 
so I have a force of 10 N pulling to the left?
 
Considering the first cart:

Let us keep the FBD in mind.

The first cart has 30N to the left and the tension T to the right. Hence Fnet in the direction of the acceleration is (30 - T).

using Fnet = ma (in the direction of the acceleration)
30 - T = 5 x 2

etc.

The same may be done for the other two carts. Always keep the FBD in mind.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
1K
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 52 ·
2
Replies
52
Views
3K
Replies
15
Views
4K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 33 ·
2
Replies
33
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 13 ·
Replies
13
Views
5K