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gasapple
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Wondering if someone could help me with my last set problem? I've been looking at it for quite some time now. The incoming force on the block is confusing me...as a result, I'm not sure how to set it up. I've worked it with the force parallel to the incline as well as perpendicular - but not as shown in this diagram (attached) - the force on block is horizontal to ground.
A 2.00-kg block is held in equilibrium on an incline of angle 60.0 degrees by a horizontal force applied in the direction shown. If the coefficient of static friction between block and incline is 0.300, determine (a) the minimum value of the applied Force and (b) the normal force exerted by the incline on the block.
Would you still set this problem up as F = mg X sin (angle) - (static friction coeff.) X mg X cos (angle)?
And how do you solve part (b)?
Also, I assume I would need to convert the 2 kg into Newtons - by multiplying 2.0 kg X 9.807 = 19.614N?
Any help would be appreciated.
Thanks in advance
A 2.00-kg block is held in equilibrium on an incline of angle 60.0 degrees by a horizontal force applied in the direction shown. If the coefficient of static friction between block and incline is 0.300, determine (a) the minimum value of the applied Force and (b) the normal force exerted by the incline on the block.
Would you still set this problem up as F = mg X sin (angle) - (static friction coeff.) X mg X cos (angle)?
And how do you solve part (b)?
Also, I assume I would need to convert the 2 kg into Newtons - by multiplying 2.0 kg X 9.807 = 19.614N?
Any help would be appreciated.
Thanks in advance