Vector addition and the force applied to the shaft of the pulley

[vldst]

Homework Statement

Hello, I am currently studying mathematics and physics on my own, and I ran into a type of problem in physics that is a bit unclear to me.

Say we have pulley at the top of an inclined plane with a base angle of alpha and the top angle is alpha.

Homework Equations

I assume that iregardless if the two masses m1 and m2 are accelerating or not, the tension to the left side of the pulley will be equal to the tension on the right side of it. Now, the force applied to the pulley's shaft will be equal to the sum of the two tension vectors, which are equal in magnitude but they have different directions. If we know the angle alpha we can also find the top angle beta which is equal to ( 180 - 90 - alpha ) degrees.

3. The Attempt at a Solution

I understand the law of cosines, but my confusion comes from the two versions of it : the first case c^2 = a^2 + b^2 - 2cos(angle)a * b and the second case c^2 = a^2 + b^2 + 2cos(angle)a*b where a and b are the vectors to be added and c is the resultant vector. After browsing the web, I've found the explanation that the difference comes from whether we're adding the vectors using the parallelogram rule or the triangle rule. It seems that we use the +2cos(angle)a*b for the parallelogram rule.

Could someone tell me if my assumptions were correct and what are the proofs. Thanks, I hope I've been comprehensible enough, english is not my first language.

 

[BvU]

Hello vldst,

Hats off for your brave enterprise. It's fun !

Re cosine rule: there is only one cosine rule, the one with the minus sign.

The other one is simply a calculation of the length of a sum vector $\vec c = \vec a + \vec b$.
In addition to what is told here:
$$\|\vec c\|^2 \equiv \vec c\cdot\vec c = ( \vec a + \vec b ) \cdot ( \vec a + \vec b ) = \vec a \cdot \vec a + \vec b \cdot \vec b + 2 \, \vec a \cdot \vec b =
\|\vec a\|^2 + \|\vec b\|^2 + 2\, \|\vec a \| \| \vec b \| \cos\theta $$
For yourself you can draw two vectors and add them using the definition of sine and cosine and Pythagoras.

 

[tnich]

Homework Statement

Hello, I am currently studying mathematics and physics on my own, and I ran into a type of problem in physics that is a bit unclear to me.

Say we have pulley at the top of an inclined plane with a base angle of alpha and the top angle is alpha.

Homework Equations

I assume that iregardless if the two masses m1 and m2 are accelerating or not, the tension to the left side of the pulley will be equal to the tension on the right side of it. Now, the force applied to the pulley's shaft will be equal to the sum of the two tension vectors, which are equal in magnitude but they have different directions. If we know the angle alpha we can also find the top angle beta which is equal to ( 180 - 90 - alpha ) degrees.
View attachment 225093
3. The Attempt at a Solution

I understand the law of cosines, but my confusion comes from the two versions of it : the first case c^2 = a^2 + b^2 - 2cos(angle)a * b and the second case c^2 = a^2 + b^2 + 2cos(angle)a*b where a and b are the vectors to be added and c is the resultant vector. After browsing the web, I've found the explanation that the difference comes from whether we're adding the vectors using the parallelogram rule or the triangle rule. It seems that we use the +2cos(angle)a*b for the parallelogram rule.

Could someone tell me if my assumptions were correct and what are the proofs. Thanks, I hope I've been comprehensible enough, english is not my first language.

I see one thing that is not clear to me. You show triangle as a right triangle, but you have drawn the base at a slant. Did you intend to do that?

You can check whether your assumptions about the law of cosines make sense or not by plugging in values of β. Try β = 0 and β = π and see which version of the law of cosines gives you the result you want.

 

[Dr Dr news]

Since this inclined plane has one base angle = 90°, you need not invoke the law of cosines which considers oblique triangles. The one assumption you can make without qualification is that a(m1) = a(m2), unless the connecting cable stretches.

 

[haruspex]

the parallelogram rule or the triangle rule

The difference is that you use different angles.
Draw a vector pointing to the right from O to A, and from A a second vector pointing up and right at angle θ to the horizontal to point B.
The angle between the vectors is θ.
Form the resultant as a vector from O to B.
The angle in the triangle is π-θ. The cosine rule says the length of the resultant is given by c²=a²+b²-2ab cos(π-θ)=a²+b²+2ab cos(θ).