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IMDB rating: 4.70 Plot: “Slip” is the story of Sarah, a young woman who witnessed her mother’s murder as a child. Now grown, Sarah is haunted by her mother’s memory, her restless soul, opening a gateway for the spirits of the recently departed. Sarah is compelled to do their bidding so their souls may rest in peace. One of these souls leads her to Cal, a young car thief who is one car away from going legit. Unfortunately, Cal is wrongfully accused of stealing from his partners and now they want him dead.Sarah, drawn by the spirit of Cal’s murdered nephew, becomes an avenging angel, awakening a dark spirit of her past while risking all to save a total stranger. Now in a race for their lives as well as their souls, Cal and Sarah soon discover that there are far more fearsome things than men with guns. |
Available versions:
DivX Version (Normal Quality), iPod/iPhone Version
Actors: Thriller
Physics help please??!!?
wo blocks, stacked one on top of the other can move without friction on the horizontal surface as show below. The surface between the two blocks is rough, however, with a coefficient of static friction equal to 0.47. If a horizontal force, F, is applied to the 5-kg bottom block, what is the maximum value F can have before the 2 kg block begins to slip?
The maximum acceleration the two blocks can have before the 2 kg block begins to slip is given by a/g=0.47. So a = 9.81 x 0.47= 4.61 m/s^2. So the maximum force must be 4.61 x 7= 32.3 newtons.
zee_prime | Nov 20, 2009
The maximum horizontal force that friction can apply to the top block is equal to the normal force times the coefficient. Since the normal force in this case is equal to the gravitational force:
Fhmax = m x g x mu = 2 x 9.8 x .47 = 9.2 N.
Divide out the m to get the maximum acceleration,
Amax = 9.8 x .47 = 4.6 m/s^2.
If the force on the bottom block is less than the amount needed to accelerate the 7 kg (5 and 2 kg blocks together) by that amount, friction will carry the top block along; if it is more, then the top block will slip.
F < 7 kg x 4.6 m/s^2 = 32.2 N.
At that force, the 5 kg block will apply a 9.2 N force forward to the 2 kg block, accelerating it by 4.6 m/s^2. The 2 kg block applies a 9.2 N force backwards (equal and opposite) on the 5 kg block, giving a net force of 32N – 9.2N = 23N, giving it an equal acceleration.
bart6500@pacbell.net | Nov 20, 2009
The maximum value of force F before the 2kg block begins to slip is
F=umg = 0.47*2*9.8 =0.932 N.
lahari s | Nov 20, 2009
The questions asks you to find the static friction (max) of the block at the top.
Fs max = coefiiceint of static friction * normal force
= .47 * (5*9.8)
= 23.03 F
Justin | Nov 20, 2009
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