Question

A meter stick whose mass is 0.200 kg is supported at the zero cm mark by a knife edge and a force F at the 100 cm point. A mass of 700 grams is attached to the stick at 40 cm mark. Find the magnitude of N and F in Newton's.

Answer 1

Start by drawing a picture. You have a meter stick of supposedly uniform mass, so the force due to gravity should be in the geometric center, pointing down. We know on the left end we have an upwards force due to the knife, and at the 40 cm mark a weight. Let's pick the knife as our pivot to eliminate that unknown force. Visualizing the gravitational force on the meter stick and the weight, we conclude that the fore F must be upward.

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P W C F

where P is the pivot, W the weight, C the center of mass of the meter stick, and F the force.

The system is in equilibrium so CCW forces must equal the CW forces. Torque is the cross product of r and F, where r is the moment arm, but since the forces are all perpendicular to the meter stick this simplifies nicely to force * distance from pivot.

CCW = CW
F(m)d(m) + F(c)d(c) = Fd(F)
(.7g)(.4) + (.2g)(.5) = F(1)
F = 3.72N [positive y-direction]

Answer 2

Would the magnitude of N also be 3.72 N?

Answer 3

What's N supposed to be? The knife?

If so, you would have to pick a different pivot and repeat the process for it.

Answer 4

N be going up on the knife edge with the 0.200 kg mass.

Answer 5

Yeah. Set up your static equilibrium equations relative to another pivot. I would recommend doing it at the 100 cm mark, where F is, but it doesn't really matter so long as it's not at the 0 cm mark.

Answer 6

Thank you so much!