...it is known to be at rest. (In particular the mass on the incline of 25 deg above the horizontal is 20kg while the haning mass is 10 kg) Apparently the correct answer would be the direction towards to bottom of the ramp but my inclination is to have it opposite, so pointing up the ramp seeing that 20 kg > 10 kg. Any help/explanation would be much appreciated.
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The force on the 20kg mass along the slope and down, from its own weight is
mg sin(angle) = 20kg* 9.8 m/ s^2 * sin(25)= 82.8 N
The force along the slope but upward - from the hanging mass- is
M g = 10 kg * 9.8 m/s^2 = 98 N
Hence the frictional force must be directed downward along the slope (15.2 N)
mg sin(angle) = 20kg* 9.8 m/ s^2 * sin(25)= 82.8 N
The force along the slope but upward - from the hanging mass- is
M g = 10 kg * 9.8 m/s^2 = 98 N
Hence the frictional force must be directed downward along the slope (15.2 N)