a box of books weighing 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exerted downward at an angle of 35.2 degrees below the horizontal. Find ,uk between the box and the floor...?
2007-12-18 05:22:30 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The normal force is
325+425*sin(35.2)
therefore
(325+425*sin(35.2))*µk=425*cos(35.2)
µk=0.61
j
2007-12-18 05:35:31 · answer #1 · answered by odu83 7 · 0⤊ 1⤋
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The weight of box of books = w = 325 N
The box is pushed with a force of 425 N exerted downward at an angle of 35.2 degrees below the horizontal.
Force that pushes the box = F =425 N
Angle which the force makes with horizontal =O =35.2 degree
Horizontal component of force = FcosO=425*0.8171=347.29
Horizontal component of force = 347.29 N
Vertical downward component of force = FsinO=425*0.5764=244.98
Vertical downward component of force = 244.98 N
The box moves at a constant velocity across the floor .
Resultant force on the box is zero
Normal reaction = R= Weight + vertical downward component of force
Normal reaction = R= 325+244.98=569.98 N
The force of friction balances the horizontal component of force because the box moves at a constant velocity across the floor .
The force of friction =mu*Normal reaction=mu*569.98
The force of friction = the horizontal component of force
mu*569.98 = 347.29
mu =347.29 /569.98=0.6092
The coefficient of kinetic friction between box and floor is 0.6092
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2007-12-18 06:00:46 · answer #2 · answered by ukmudgal 6 · 1⤊ 1⤋