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Comments on Why used $\cos\theta$ for $\text{y}$ axis or, gravitational force?

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Why used $\cos\theta$ for $\text{y}$ axis or, gravitational force?

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figure 3.2figure 3.1
Mass M1 is held on a plane with inclination angle θ, and mass M2 hangs over the side. The two masses are connected by a massless string which runs over a massless pulley (see Fig. 3.1). The coefficient of kinetic friction between M1 and the plane is µ. M1 is released from rest. Assuming that M2 is sufficiently large so that M1 gets pulled up the plane, what is the acceleration of the masses? What is the tension in the string?

Then, they were writing force of that figure.

$$T-f-M_1g\sin \theta = M_1a$$ $$N-M_1g\cos \theta=0$$ $$M_2g-T=M_2a$$

In the second equation they wrote that $$M_1g\cos \theta$$

Usually, $\cos$ is used when we think of $\text{x}$ axis. Since, $$\cos \theta=\frac{\color{blue}\text{base}}{\text{hypotenuse}}$$ But, gravitational force is forever through $\text{y}$ axis. Although, why they used $\cos\theta$ for gravitational force.

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1 comment thread

Doesn't this question actually belong on Q&A? (3 comments)
Doesn't this question actually belong on Q&A?
celtschk‭ wrote over 3 years ago

Doesn't this question actually belong on Q&A?

deleted user wrote over 3 years ago

celtschk‭ It is well-fitted in Q&A but, I found the problem in "Problems" that's why I asked it here. If you think that Q&A is better than Problems for the question than, you can flag for mod attention. They will move to Q&A

celtschk‭ wrote over 3 years ago

The reason why I think it doesn't belong to the Problems category is that it doesn't contain your attempt to solve it, but a question about a solution given in a text book. But then, my understanding on what this category is for might be wrong, which is why I only commented with a question.