frosty bay
Can't figure out:
I know -
The magnitudes are:
Pedal down = 10 N
Shafts length = 20cm = 0.2m
Sine Theta Angle Between:
Shafts position = pi / 6 = 30 degrees
The torque's magnitude of the bike = the perpendicular vector's magnitude on the point where the shaft and pedal are involved
(10)(0.2)sin(30) = 1 Nm
What's the error??
ebon ploverBOT
hard pewter
Are you sure that the component of the force isn't 10cos30?
frosty bay
Are you sure that the component of the force isn't 10cos30?
How?
I entered 5sqrt(3) it doesn't work
hard pewter
component of the force is still multiplied by distance to get torque, so I am suggesting sqrt{3} Nm
When you resolve the vertical force into components perpendicular to and parallel to the shaft, the perpendicular component is adjacent to the 30 degree angle, hence cosine
Assuming that I am reasoning this correctly
brittle ember
Yeah
You need to take into account the force which is going to be applied from a completely vertical force
frosty bay
@brittle ember @hard pewter Can you expand more fully, I don't understand the logic fully
And why the other approach was wrong
brittle ember
The torque force is perpendicular
Which would be tangential to the circular path
So a completely vertical force excerted on the shaft would be the only ocassion where the torque force is equal to length times the force
Is it alright if i send a diagram to show how it works?
frosty bay
Is it alright if i send a diagram to show how it works?
? @brittle ember
brittle ember
Right
With this value for the force (5 sqrt(3) not 10 (i used the wrong value its 10 not 20)) applied we just need to multiply it by the lenth radius to get the torque
$t= F\times L \newline$
$t=t\sqrt{3} \times 0.2 \newline$
$t=\sqrt{3} Nm$
copper jackalBOT
imdeadinside4630
brittle ember
@frosty bay