Solving for applied force to push an object up a ramp.

[agentnnc]

A box, mass 100kg, must be pushed onto a table using an inclined plane 1.5 m high. The plane is 4 m long. If the pleane offers a frictional force equivalent to 10% the weight of the box, what force must be applied to move the box?


What I tried:
sin^-1(4/1.5) = 22.6
100 * 9.8 = 980 N (fgrav)
980cos22.6=904.77N (Fperpendicular)
Ffrict=10%980=98N

That's where I get stuck. I'm not sure if I'm even doing it right (However, I know the angle is).

Please help. My exam is tomorrow.

 

[inky]

A box, mass 100kg, must be pushed onto a table using an inclined plane 1.5 m high. The plane is 4 m long. If the pleane offers a frictional force equivalent to 10% the weight of the box, what force must be applied to move the box?


What I tried:
sin^-1(4/1.5) = 22.6
100 * 9.8 = 980 N (fgrav)
980cos22.6=904.77N (Fperpendicular)
Ffrict=10%980=98N

That's where I get stuck. I'm not sure if I'm even doing it right (However, I know the angle is).

Please help. My exam is tomorrow.

angle (theta)=sin^-1(1.5/4)=22.02 deg. (You take sin^-1 (4/1.5))

Use summation F(x)=0
friction and mg sin (theta) direction is downward . Applied force dirction is upward.Yes your friction is 98 N. No need to consider Y direction here.