Why is tension on a slope equal to the sum of the frictional force and the component of the weight? I don't really understand why...

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anonymous · Answer

For example:
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anonymous · Answer

It depends on what is happening when the object is tied to the rope. As you can see, its weight is not balanced (it's not entirely perpendicular to the incline, as expected) and there is a component of it parallel to the slope: $w\sin(\theta)$.

One way to understand tension is by looking at the object being dragged. By doing so, we know that the tension is:

Tension = Force being overcomed + Force exerted to make the object move with an acceleration.

If the object is neither sliding down nor being pulled up ($v=0$) then the tension is just the force being overcomed (otherwise it will slide down), which is:

$\large{T = w\sin(\theta)}$

If the object is being pulled up the slope (as it can be understood by what you were told). It can happen 2 things:

1) The object is being pulled with $a=0$, the force being overcomed is $w\sin(\theta)$ as before, but know the kinetic friction is present during the path travelled:

$\large{T = w\sin(\theta) + f_k}$

2) The object is being pulled with $a > 0$, then the forces are the same as in 1) but the motor is exerting an extra force, because the object of mass $m$ is being given an acceleration $a$:

$\large{T = w\sin(\theta) + f_k + ma}$

I guess what you were told is referring to situation 1).

In the case that you were wondering what is the other approach by which this types of tension problem can be addressed, it's not by looking at the object being dragged but at the thing that is pulling it. In that case:

Tension = Total Force exerted by that thing - Net Force exerted by that thing