Finding Tension and Acceleration

In summary, a block of mass 6.0 kg on a rough surface with a coefficient of kinetic friction of 0.1 and tilted at an angle of 30 degrees is connected by a string to another block of mass 2.0 kg on a frictionless surface. A horizontal force of 3.0 N is acting on the first block, causing an acceleration of -0.31 N. The normal force acting on the second block is approximately 49.19 N and the magnitude of acceleration for the second block is approximately 0.31 N.
  • #1
B3NR4Y
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Homework Statement


In figure, block 1 of mass m1 = 2.0 kg and block 2 of mass
m2 = 6.0 kg are connected by a massless string over a massless and
frictionless pulley and are initially held in place. Block 2 is on a rough
surface, with coefficient of kinetic friction of 0.1 and tilted at θ = 30o
while a horizontal force F of magnitude 3.0 N acts on it. Block 1 is on a frictionless surface.
Find the normal force acting on block 2.
Find the magnitude of acceleration of block 2.

Homework Equations


F= ma
Fg = -mg
Fg, y =-mg*cos θ
Fg, x =mg*sin θ

The Attempt at a Solution


I drew a free body diagram of each individual block, for block one I found it has a force in the positive x-direction, the force is the force of tension (therefore it will move in the direction of tension) and the force of gravity on block one is canceled by the normal force so F T = m_ 1 * a. I solved for a, and got a = FT/m1. My reason for doing this is I reasoned the acceleration is the same for both blocks, which I may be wrong in assuming this. I solved for the x and y component of the horizontal force that acts on block 2, because my x-axis runs up to down and the up direction is positive for the y axis.

To solve for the normal force, I found the y-component of the force from gravity, which is -m2 g cos 30o. I then said Y-component of gravity + Normal Force + Y-component of horizontal force = 0 and solved for Normal force. I got ~49.19 N

Then to find tension I added all the forces in the x-direction and solved for tension.

Tension + Friction Force + Force of Gravity in the X-direction + X-component of the horizontal force = m2 a
I solved for a earlier, and got FT/m1

I did the algebra and found the force of tension to be -0.31 N to be the force of tension, which has to be wrong because when I sum the forces I get a negative number.

Basically I would like to know how to solve for tension and acceleration when all you're given is the coefficient of friction, a force, the angle of the incline, and the two masses. In class we did an example where we had to solve for the tension, but we were given all this information and the acceleration of the object.
 
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  • #2
This is hard to follow without the picture.

because my x-axis runs up to down and the up direction is positive for the y axis
is really a non-sequitur

It seems block 1 is on a horizontal plane ?
Is the string vertical from pulley to block 2 ?
The 3 N is working in which direction ?
How far does block 2 move when block 1 moves a small distance s ?
 
  • #3
BvU said:
This is hard to follow without the picture.

is really a non-sequitur

It seems block 1 is on a horizontal plane ?
Is the string vertical from pulley to block 2 ?
The 3 N is working in which direction ?
How far does block 2 move when block 1 moves a small distance s ?
Silly me, I'm stupid I knew I needed a picture... I'm on my phone currently but will post one
 
  • #4
5cgbE0V.png

Sorry friend, here it is
 
  • #5
I got the same result as you for the normal force on m2. Please flesh out algebraically the force balance on m2 in the direction parallel to the incline. Writing out in terms of words is not useful to us for identifying where you encountered your problem. Thanks.

Chet
 
  • #6
Picture sure helps! You don't want to know what I concocted from the description only :-)
Tension + Friction Force + Force of Gravity in the X-direction + X-component of the horizontal force = m2 a
looks good, so I suspect a mishit on the calculator (*). As Chet says: show the terms (with signs!)

(The more so because I didn't get 49.2 at all: ## -(-m_2g\cos{\pi\over6} + F\sin{\pi\over6}) =49.4 \ ## if I use g = 9.8 m/s2 :-) )
 
  • #7
BvU said:
Picture sure helps! You don't want to know what I concocted from the description only :)
looks good, so I suspect a mishit on the calculator (*). As Chet says: show the terms (with signs!)

(The more so because I didn't get 49.2 at all: ## -(-m_2g\cos{\pi\over6} + F\sin{\pi\over6}) =49.4 \ ## if I use g = 9.8 m/s2 :) )
It must have been a mishit into my calculator or I just made up some dark energy numbers (joking), I do practice problems on a whiteboard so I don't waste paper and I got the wrong answer on the whiteboard. Today I was sitting in the library so I decided to rework the problem on paper, and I got what I assume is the right answer (I don't have the solution, it's a midterm from last year). I am not sure what I did to get the numbers I was getting :S thanks for your helps/potential helps!
 

FAQ: Finding Tension and Acceleration

What is tension and how is it related to acceleration?

Tension is a force that occurs when an object is pulled or stretched. It is related to acceleration because it is one of the forces that can cause an object to accelerate. When an object experiences tension, it will either speed up or slow down depending on the direction of the force.

How do you calculate tension and acceleration?

To calculate tension, you need to know the mass of the object and the acceleration it is experiencing. The formula for tension is T = ma, where T is tension, m is mass, and a is acceleration. Acceleration can be calculated using the formula a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity, and t is the time interval.

What are some common real-world examples of tension and acceleration?

Some common examples of tension and acceleration include a car accelerating on a straight road, a pendulum swinging back and forth, and a person jumping on a trampoline. In these examples, the tension force is caused by the acceleration of the object or person.

How does the angle of the rope or string affect the tension and acceleration?

The angle of the rope or string can affect the tension and acceleration in a few ways. If the rope is horizontal, the tension will be equal to the weight of the object being pulled. If the rope is at an angle, the tension will be greater than the weight of the object. This increased tension can cause the object to accelerate at a faster rate.

What factors can cause tension and acceleration to change?

The factors that can cause tension and acceleration to change include the mass of the object, the force being applied, and the angle of the rope or string. Additionally, friction and air resistance can also affect tension and acceleration by opposing the motion of the object and reducing the net force acting on it.

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