Dynamics Question: Acceleration and Tension Calculation

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In summary, the unknown minimum horizontal force required to hold the blocks in equilibrium is less than 40 N.
  • #1
ME_student
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Homework Statement


I feel like I've spent way too much time on this problem... So the given information is on the image along with all my work. I am finding the acceleration of the block and the tension in the cable. I know R is going to be the same in all the cords because the pulleys are mass less and we are neglecting friction the pulleys. I first analyzed which way the forces of friction was were applied at each box then I sum of forces in x and y for both blocks but I need to first see if the block is moving. By doing so I used the coefficient of static friction and summed the forces to zero.

So when I solved for P to determine if the block is moving I got -24.55N which is less than 40N, therefore the block is sliding because P <40N.

But the answer to P is -19.04N all my equation are riGht. I verified with the solution key and they match exactly the same but I can't get -19.04N

Homework Equations


On imagr

The Attempt at a Solution


On image
 

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  • #2
Images are sideways and all but unreadable -- not enough definition/resolution.

Why don't you draw up a larger image and describe the problem and your work in text, in the post (as the forum rules require!)?
 
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  • #3
Sorry, it's difficult to read your work. It might help if you can take the picture directly overhead instead of an oblique angle. And also turn the page 90 degrees.
 
  • #4
You'll never make a good spy! :wink:
 
  • #5
Well I manage to figure out part A and part B of the question... Still let's discuss why my P force is wrong. Let me retake a few pics.

Better?
 

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  • #6
gneill said:
Images are sideways and all but unreadable -- not enough definition/resolution.

Why don't you draw up a larger image and describe the problem and your work in text, in the post (as the forum rules require!)?
Okay I can do that... Next time. :smile:
 
  • #7
Better but I'll have to rotate my screen to read.
 
  • #8
gneill said:
Images are sideways and all but unreadable -- not enough definition/resolution.

Why don't you draw up a larger image and describe the problem and your work in text, in the post (as the forum rules require!)?
Not trying to be a smarty pants but that's why you rotate the image and zoom into identify the images and calculatuons.
paisiello2 said:
Better but I'll have to rotate my screen to read.
Let me try to rotate the images...
 
  • #9
How's that?
 

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  • #10
ME_student said:
Not trying to be a smarty pants but that's why you rotate the image and zoom into identify the images and calculatuons.
Hey, you're here for FREE help from volunteer experts. Doing what you can to make their life easier to help you would seem to be the smart play.

That's one reason why the rules require the template format and discourage image-only problem and solution statements. Helpers should not have to work hard to understand the question being posed -- a simple read of the a text problem statement with perhaps a visible embedded image to clarify should be sufficient to grok the situation.

I know that many helpers will simply pass over posts that require too much effort to get to the problem. They prefer to go on to posts where the member has made the effort up front.
 
  • #11
Starting with block A

Sum of the forces in x direction, right being plus (+)
Equation 1: -T-mu_s*FN1-MU_s*FN2+W_A*sin (25)+Pcos (25)=0
Zero because the block is static.

Sum of the forces in y direction, up being plus (+)
Equation 2: FN1-FN2-W_A*cos (25)+P*cos (25)=0 Now looking at block B

Sum of the forces in y direction
Equation 3: FN2-W_B*cos (25)=0
Zero because we are trying to determine if the blocks will slide.

Sum of the forces in x direction
Equation 4: -T+Fm2+W_b*sin (25)=0

We can now solve for FN2 and T

So once I solve for T and FN2 I use equation 1 and 2 to solve for P
 

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  • #12
Your free body diagrams look good but what exactly is P now? This is some unknown minimum horizontal force required to hold the whole thing in equilibrium?
 

FAQ: Dynamics Question: Acceleration and Tension Calculation

What is acceleration and how is it calculated?

Acceleration is the rate of change of an object's velocity over time. It is calculated by dividing the change in velocity by the change in time, or by using the equation a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

How does acceleration affect an object's motion?

Acceleration affects an object's motion by changing its velocity. If an object is accelerating, its velocity will either increase or decrease depending on the direction of the acceleration. This means that the object's speed and/or direction will change over time.

What is tension and how is it calculated?

Tension is the force that is transmitted through a string, rope, cable, or other similar object when it is pulled tight by forces acting on both ends. It is calculated by using the equation T = m * a, where T is tension, m is the mass of the object, and a is the acceleration of the object.

How does tension affect objects in motion?

Tension affects objects in motion by providing the force needed to accelerate them. For example, in a pulley system, tension in the rope is what allows objects to move up or down. In a pendulum, tension in the string is what keeps the pendulum moving back and forth. Without tension, the objects would not be able to move.

How can acceleration and tension be calculated in real-world scenarios?

Acceleration and tension can be calculated in real-world scenarios by using the equations a = (vf - vi) / t and T = m * a, respectively. In order to use these equations, you will need to know the values of the object's velocity, time, and mass. Real-world scenarios may also involve other factors such as friction and air resistance, which can affect the accuracy of the calculations.

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