physics help w/ question…?

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…?

2 Answers

  • odu83
    1 month ago

    The normal force is

    325+425*sin(35.2)

    therefore

    (325+425*sin(35.2))*µk=425*cos(35.2)

    µk=0.61

    j

  • ukmudgal
    1 month ago

    _______________________________________

    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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