Asked by Serina
A 10 kg block is pressed against a vertical wall and held in place by a force F of 180 N, 37° above the
horizontal. The coefficient of static friction μS=0.20. Find the magnitude and direction of the frictional force.
The answer is 10.3 N downward but I don't know how to approach this.
I know friction force is static coefficient x Normal force. Normal force is perpendicular to the surface. Shouldn't the friction force be 0.2 x 180cos37?
horizontal. The coefficient of static friction μS=0.20. Find the magnitude and direction of the frictional force.
The answer is 10.3 N downward but I don't know how to approach this.
I know friction force is static coefficient x Normal force. Normal force is perpendicular to the surface. Shouldn't the friction force be 0.2 x 180cos37?
Answers
Answered by
Anonymous
180 sin 37 = 108 N up from force
10 * 9.81 = 98.1 N down from weight
difference is to be made up by friction
108 - 98 = 10 N friction down
now is that less than the friction max allowed?
that is your .2*180 cos 37 = 28.7
so we have much more friction available than we need
10 * 9.81 = 98.1 N down from weight
difference is to be made up by friction
108 - 98 = 10 N friction down
now is that less than the friction max allowed?
that is your .2*180 cos 37 = 28.7
so we have much more friction available than we need
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