# How can a rope's tension be at a maximum as a body is about to move up a slope in the rope's direction? [closed]

Note: I do NOT need this exercise's answer. I have no problem working the math out and noticing the truth of my title.

A box of mass 0.5 kg is placed on a plane which is inclined at an angle of 40 degrees to the horizontal. The coefficient of friction between the box and the plane is 1/5. The box is kept in equilibrium by a light string which lies in a vertical plane containing a line of greatest slope of the plane. The string makes an angle of 20 degrees with the plane, as shown in the diagram. The box is in limiting equilibrium and may be modelled as a particle. The tension in the string is $$t$$ newtons.

Find the range of possible values of $$t$$.

How can a rope's tension be at a maximum as a body is about to move up a slope in the rope's direction?

Perhaps my confusion is due to not really understanding how tension works. I know there are complicated tension based interactions both internal and external acting on a body.

In earlier question's tension in a rope was seen as working in the opposite direction to a force acting on a weight held by said rope. In this case it is not quite the opposite direction, but I can not say that intuitively I understand why the tension would be at a maximum as the body is about to move in the general direction of the rope.

## closed as off-topic by Aaron Stevens, Yashas, Jon Custer, Dvij Mankad, M. EnnsMay 16 at 12:39

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When you say that the tension is "at its maximum value", you mean "at the maximum value that still allows for limiting equilibrium". At some point in your solution, you probably assumed that the acceleration of the block vanishes ($$a = 0$$); this is the assumption of limiting equilibrium. You could certainly pull on the rope with a tension greater than this critical value, but then the block would start accelerating and it wouldn't be in equilibrium any more.