A block of mass 6 kg on an incline angled at 24 degrees above the horizontal experiences a frictional force resisting its tendency to slide along the ramp. The upper limit on this frictional force is .13 of the normal force between the block and the ramp. What force must be exerted parallel to the ramp so that the block will slide at a uniform velocity up the ramp?
2007-07-03 05:24:22 · 2 answers · asked by benzene boy 1 in Science & Mathematics ➔ Physics
In order to achieve this, you must exactly overcome the frictional force as well as the component of gravity acting parallel to the ramp. The component of the gravitational force is mg*sin(t) and the frictional force is umg*cos(t), where m is the mass of the block (6 kg), g is the acceleration due to gravity (9.8 m/s^2), t is the angle of the incline (24 degrees), and u is the coefficient of friction (0.13). So the force must be equal to mg*sin(t) + umg*cos(t), which you can calculate.
It should be noted that the Greek letters mu and theta would normally be used rather than the Latin letters u and t, and that the coefficient of friction should be the kinetic rather than static coefficient, while the coefficient given here is not qualified.
2007-07-03 05:29:47 · answer #1 · answered by DavidK93 7 · 0⤊ 0⤋
The sum of the forces must be zero:
down-slope gravity: mg sin theta
friction (down slope): mu mg cos theta
force exerted (up slope): F
So F = mg (sin theta + mu cos theta)
They give you m, mu, and theta. You know g. Plugnchug.
2007-07-03 12:30:10 · answer #2 · answered by Anonymous · 0⤊ 0⤋