Block on an inclined plane/ tension / friction coefficient

A rectangular block of mass 30kg rests on a plane inclined at 30 degrees to the horizontal. The block is held in equilibrium by a string, which is parallel to the plane.
1 - Calculate tension in string if plane is smooth.
2 - Given that the coefficient of friction between the block and the plane is 0.6, calculate the tension in the string if the block is just on the point of moving up the plane.
~many thanks
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1 - Calculate tension in string if plane is smooth.
Since there is no friction, the tension in the string must equal the force parallel
Tension = 30 * 9.8 * sin 30˚
2 - Given that the coefficient of friction between the block and the plane is 0.6, calculate the tension in the string if the block is just on the point of moving up the plane.
Since there is friction, the tension = Force parallel – Friction force.
Tension = (30 * 9.8 * sin 30˚) – (0.6 * 30 * 9.8 * cos 30˚) = 147 – 152.8 = -5.8 N
This is not possible, because the friction cannot push the block up the incline.
When the friction force is greater than the force parallel, the tension is equal to 0 N
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Nevermind, got the answer and working, but idk how to delete this :/

Quote:

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Originally Posted by patrickmanning

A rectangular block of mass 30kg rests on a plane inclined at 30 degrees to the horizontal. The block is held in equilibrium by a string, which is parallel to the plane.
1 - Calculate tension in string if plane is smooth.
2 - Given that the coefficient of friction between the block and the plane is 0.6, calculate the tension in the string if the block is just on the point of moving up the plane.
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1 - Calculate tension in string if plane is smooth.
Since there is no friction, the tension in the string must equal the force parallel
Tension = 30 * 9.8 * sin 30˚
2 - Given that the coefficient of friction between the block and the plane is 0.6, calculate the tension in the string if the block is just on the point of moving up the plane.
Since there is friction, the tension = Force parallel – Friction force.
Tension = (30 * 9.8 * sin 30˚) – (0.6 * 30 * 9.8 * cos 30˚)

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and this equation would be incorrect ...

Oh no! Well could you help me out? I got that from another site.

which direction does the max static friction force act in this situation? (note what I bolded in your post response)

Sorry, no idea, man. That's exactly all the question gave.
*Edit: Although I'm guessing the static friction tends to act DOWN the plane as it should oppose the block which is on the point of slipping UP the plane.
Does the final answer end up positive or negative? I got T = 300N