On her way in to work one morning, an evil chemistry professor notices an 80 kg

Question

On her way in to work one morning, an evil chemistry professor notices an 80 kg bag of cats dangling next to the science building, hanging by a (massless) thread. The chemistry professor sees the thread is running over a massless, frictionless pulley mounted to the edge of the horizontal roof. She pulls a periscope out of her back pocket and uses it to determine that the massless rope is attached to a concrete block sitting on the roof. If the mass of the concrete block is 250 kg and the coefficient of static friction between the roof and the concrete block is 0.58, what is the minimu, force the evil professor must apply to the bag in order to overcome static friction and lower the cats to safety? Include two free-body diagrams.

Explanation / Answer

Basically, you need the downward force exerted on the bag of cats to be greater than the force of friction on the block, so set them equal to each other.

That is: F(N)block = F(N)cats

mue*m(block)*g = m(bag)*g + F

So the force the professor would need to exert is:

F = mue*m(block)*g - (m(bag)*g)
F = 0.58*250*9.8 - (80*9.8)

F = 637 N (640 when rounded to sig. figs.)

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