Using the sin function for a problem with a frictionless pulley and an incline

In summary, the sin function can be used to solve problems involving a frictionless pulley and an inclined surface. By analyzing the forces and angles involved, the sin function can determine the acceleration and tension of the system. This can be applied to various scenarios, such as calculating the speed of an object sliding down an incline with a pulley attached. The use of the sin function allows for a more accurate and efficient solution to these types of problems.
  • #1
sentimentaltrooper
6
1
Homework Statement
A system comprising blocks, a light frictionless pulley, a frictionless incline, and connecting ropes is shown in the figure. The 9.0-kg block accelerates downward when the system is released from rest. The tension in the rope connecting the 6.0 kg block and the 4.0 kg block is closest to
Relevant Equations
[itex](m_2 + m_1)a - Ma = Mg - m_2 gsin\theta - m_1 gsin\theta[/itex]
image?k=fc77d489804f17f65c75a97e435e4076.jpg

To find the tension in the rope connecting 6.0 kg block and 4.0 kg block we do
6.0 kg = m1, 4.0 kg = m2, 9.0 kg = M
[itex](m_2 + m_1)a - Ma = Mg - m_2 gsin\theta - m_1 gsin\theta[/itex]

Why do we use sin in these equations and not cos?
 
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  • #2
sentimentaltrooper said:
Why do we use sin in these equations and not cos?
Please explain why you think it would be cos.
 
  • #3
Another student that hasn't been introduced to the basics of trigonometry before taking the course in mechanics...
Do you know how the cosine and sine of an angle of a right triangle are defined?
 
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  • #4
There are some of us (I am one) for whom seeing things in horizontal and vertical is much easier than seeing them at angles like this. But here, if you were to consider motion of the 6kg and 4kg masses horizontally and vertically, you'd be making things more complicated than they need to be. It's more efficient here to take the components parallel to and perpendicular to the slope. So, what matters here is the component of weight (of the 6kg and 4kg) parallel to the slope, because that's where the motion is, parallel to the slope. This means you need a right angled triangle which has the weight as the hypotenuse. (I usually turn the paper, or my head, to see this better.)

Anyway, try it. Find the component of the weight parallel to the slope and see whether you need the sin, cos or tan of that angle to do so.
 
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  • #5
What is the direction of the only effective forces inducing and resisting the movement?
Would the direction of the movement reverse for angles close to 90 degrees?
 
  • #6
Delta2 said:
Another student that hasn't been introduced to the basics of trigonometry before taking the course in mechanics...
Do you know how the cosine and sine of an angle of a right triangle are defined?
i know the basics of trig
soh cah toa
so sin would be... opposite over hypotenuse... but opposite length of triangle is not even defined here and neither is hypotenuse so what use is this sin ?

The component of weight parallel to slope:
we have 6 kg and 4 kg so just multiply both by 9.8 no?
(6 * 9.8) + (4 * 9.8) = 98 N
 
  • #7
sentimentaltrooper said:
The component of weight parallel to slope:
we have 6 kg and 4 kg so just multiply both by 9.8 no?
Those are the whole weights, not components in the given direction.
What you seem to be missing is how one uses trig to resolve a force (or any vector) into components.
Try https://www.physicsclassroom.com/class/vectors/Lesson-1/Vector-Resolution
 
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  • #8
haruspex said:
Those are the whole weights, not components in the given direction.
What you seem to be missing is how one uses trig to resolve a force (or any vector) into components.
Try https://www.physicsclassroom.com/class/vectors/Lesson-1/Vector-Resolution
Thank you for this website. I just used it and played through the interactive game.
So I can use square root of x component squared and y component squared to find magnitude.
And I can use head-to-tail method for adding and subtracting vectors.
But where exactly does sin come into play? And how do I know to use sin here and not cos?

I think this differentiation of concept and trig choice would allow me to understand the formula further.
 
  • #9
sentimentaltrooper said:
Thank you for this website. I just used it and played through the interactive game.
So I can use square root of x component squared and y component squared to find magnitude.
And I can use head-to-tail method for adding and subtracting vectors.
But where exactly does sin come into play? And how do I know to use sin here and not cos?

I think this differentiation of concept and trig choice would allow me to understand the formula further.
The website I linked to before is usually quite good, but looking more at that particular page it could use some improvements. This one looks better:
https://www.wikihow.com/Resolve-a-Vector-Into-Components#Calculating-Components-with-Trigonometry.
The basic rule is: given a vector magnitude X and a vector magnitude Y with an angle A between them, the component of X in the direction of Y is ##X\cos(A)##. Similarly, the component of Y in the direction of X is ##Y\cos(A)##.

Even knowing the rule, picking the wrong one of sine and cos is an easy mistake. I double check by considering an extreme case. E.g. in the problem in this thread, what if the angle were zero? The weights would not exert any tension on the rope, so which of sine and cos gives a result zero for an angle of zero?
 
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Related to Using the sin function for a problem with a frictionless pulley and an incline

1. How do you use the sin function to solve a problem with a frictionless pulley and an incline?

The sin function is used to determine the relationship between the angle of the incline and the forces acting on the system. By using the sin function, you can calculate the magnitude and direction of the forces involved, such as the tension in the rope and the normal force on the incline.

2. What is the role of the frictionless pulley in this problem?

The frictionless pulley allows for the transfer of forces between the two sides of the system without any energy being lost to friction. This allows for a more accurate calculation of the forces involved and simplifies the problem.

3. Can the sin function be used for problems with a frictionless pulley and an incline on any surface?

Yes, the sin function can be used for problems on any surface as long as the surface is smooth and there is no friction present. This allows for a more accurate and simplified calculation of the forces involved.

4. How does the angle of the incline affect the forces involved in this problem?

The angle of the incline affects the magnitude and direction of the forces involved. As the angle increases, the magnitude of the forces also increases, while the direction of the forces may change. By using the sin function, you can determine the specific relationship between the angle and the forces.

5. Are there any other mathematical functions that can be used for problems with a frictionless pulley and an incline?

Yes, there are other mathematical functions, such as the cosine and tangent functions, that can be used for problems with a frictionless pulley and an incline. However, the sin function is typically used due to its direct relationship with the angle of the incline and the forces involved.

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