You don't have to solve this problem. I recently took the MCAT and saw this problem and had no idea what to do, other than the fact that it's related to torque. The MCAT tests basic physics and this was a discrete, meaning it is designed to take the examinee roughly 1 minute to complete max. I don't care about the answer, I just made this up. How do you solve it? I think it was asking about the resultant force. REMEMBER: this question is designed to take 45 seconds - 1 minute.
http://i.imgur.com/RALe4.png
http://i.imgur.com/RALe4.png
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Take the top left corner to be your pivot point and take a look at each force individually.
For the force at the bottom left, its line of action crosses the pivot point, so no moment (torque) is produced.
The force on the bottom right has a line of action that doesn't cross the pivot point. The lever arm (perpendicular distance between line of action and pivot point), is 5 m. F*L = 50 N*m ccw
The lever arm for the top right force is 10m. F*L = 100 N*m ccw
Net torque acting on pivot point = 0 + 100 + 50 = 150 N*m counter clock wise.
For the net force, the force acting in the y axis is 10 - 10 = 0.
The net force acting in the x direction is 10N right .
So the net force, is 10N right
For the force at the bottom left, its line of action crosses the pivot point, so no moment (torque) is produced.
The force on the bottom right has a line of action that doesn't cross the pivot point. The lever arm (perpendicular distance between line of action and pivot point), is 5 m. F*L = 50 N*m ccw
The lever arm for the top right force is 10m. F*L = 100 N*m ccw
Net torque acting on pivot point = 0 + 100 + 50 = 150 N*m counter clock wise.
For the net force, the force acting in the y axis is 10 - 10 = 0.
The net force acting in the x direction is 10N right .
So the net force, is 10N right
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Hmm, bit tricky!
I don't think the 10 N downward force is relevant, because it's radial (i.e. has zero rotational component).
So in terms of what is rotating the object, you have a 10 N force acting 5 m from the centre, along an arm, and a 10 N force acting 10 m from the centre. Given that they are at right angles to the radii, I think the fact that they are in different directions is OK: they are tending to rotate in the same direction.
So the net torque would be simply (5*10)+(10*10) = 150 Nm
Happy to be corrected, but that's how I'd attack it, and it seems about the right difficulty level.
I don't think the 10 N downward force is relevant, because it's radial (i.e. has zero rotational component).
So in terms of what is rotating the object, you have a 10 N force acting 5 m from the centre, along an arm, and a 10 N force acting 10 m from the centre. Given that they are at right angles to the radii, I think the fact that they are in different directions is OK: they are tending to rotate in the same direction.
So the net torque would be simply (5*10)+(10*10) = 150 Nm
Happy to be corrected, but that's how I'd attack it, and it seems about the right difficulty level.
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The crude drawing makes it difficult...but I believe the answer is 10N to the right. You have 10N up and 10 N down...those cancel each other out. You have 10N to the right, but nothing going to the left to cancel that out...so it should be 10N to the right