A box sits on a horizontal wooden board. The coefficient of static friction between the box and the board is 0.25. You grab one end of the board and lift it up, keeping the other end of the board on the ground. What is the angle between the board and the horizontal direction when the box begins to slide down the board?
2007-09-25 12:38:20 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The force of an object on an incline is split between its Parallel and Perpendicular Force. The parallel (parallel to the incline) component is the component that counteracts the Frictional Force. Parallel Force of an object on an incline is given by the following equation:
F=m*g*sin(theta) where theta is the angle of the incline.
The Frictional Force is just the Coefficient of Static Friction times the Normal Force of the object.
Now solve for theta, when the Parallel Force equals the Frictional Force.
sin(theta)=.25*m*g/(m*g)
This simplifies to sin(theta)=.25 or Theta=14.4775 degress.
2007-09-25 13:11:59 · answer #1 · answered by erikoo7 3 · 0⤊ 0⤋
as you tip teh board up, the force of gravity on the block can be 'split' between one force parallel to the board, and one perpendicular to it. these are related to the angle of the board.
the force perpendicular to the board is the force that holds the box in place. static friction says that the sliding foce can be equal to .25 x the perpendicular force before it breaks loose.
so the perp force is equal to weight times cosine of the angle. the sliding force is equal to weight times sine of the angle. as mentioned the box will slide when sliding force is 1/4 of the perp force... so,
W x Sin theta = W x .25 x Cos theta
W cancels, Sin theta/cos theta = Tan theta = .25
from that the angle = about 14deg
2007-09-25 20:14:12 · answer #2 · answered by Piglet O 6 · 0⤊ 0⤋