A wooden block slides directly down an inclined plane at a constant velocity of 6.0m/s. how large is the coeff? - wooden block plane
a block of wood strips directly on a slope at a constant speed of 6.0 m / s. What is the coefficient of friction when the plane at an angle of 25 degrees from the horizontal?
Friday, January 8, 2010
Wooden Block Plane A Wooden Block Slides Directly Down An Inclined Plane At A Constant Velocity Of 6.0m/s. How Large Is The Coeff?
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Start with a sketch to visualize the problem. Draw an inclined plane with a block of wood. Let us now consider the forces on the block. We have the gravity and friction in a perfect balance, so that the block moves at constant velocity without acceleration.
In the diagram, draw a straight line on the wooden box. Who represents the force of gravity. This force can be divided into two components: a Council, in parallel to the slope, and another that acts perpendicularly to the slope. If you draw two lines representing the two components, drew a small triangle, with gravity, which is the hypotenuse. The shortest side of the slope (and parallel to it) is the component of gravitational force acting on the block to move down to the slope. The page that is perpendicular to the slope can be used to determine the frictional force.
The gravity is the mass of the block of wood (m) multiplied by the acceleration due to gravity (9.81 m / s ^ 2) and thereforesees
1). FG = m (9.81 m / s ^ 2), the hypotenuse of the triangle,
Well, if you refer to your plan, you will find that the gravity component parallel to the slope act must be considered
2). Fp = F (sin (25)) = m (9.81 m / s ^ 2) (sin (25))
The component that is perpendicular to the slope is given by the equation
3). FN = FG (COS (25)) = m (9.81 m / s ^ 2) (sin (25))
The normal component is used to determine the kinetic frictional force, which the normal force times the coefficient of friction. Therefore, the friction force is simply the "U", the coefficient of sliding friction, Formula 3 times, which to us that
4). Ff = UM (9.81 m / s ^ 2) (sin (25))
Now we know that keeping the block at a constant speed without acceleration, we know that the force acts parallel to the slope (Equation 2 above) is exactly equal to the frictional force (Equation 4). So, now we must
m (9.81 m / s ^ 2) (sin (25)) = UM (9.81 m / s ^ 2) (cos (25))
sin (25) = u (COS (25))
(without (25)) / (COs (25)) = u
Tan (25) = U
U = 0466
Everything is much easier than it sounds, but it is the longest way to arrive at the answer.
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