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.
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.
-------------
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]
Would the magnitude of N also be 3.72 N?
What's N supposed to be? The knife?
If so, you would have to pick a different pivot and repeat the process for it.
N be going up on the knife edge with the 0.200 kg mass.
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.
Thank you so much!
To find the magnitudes of the forces N and F, we can use the principle of torque equilibrium.
The torque exerted by a force is given by the product of the force and its perpendicular distance from the point of rotation. In this case, the point of rotation is the knife edge at the zero cm mark.
First, let's calculate the torque due to the weight of the meter stick itself. Since its mass is 0.200 kg, we can find its weight by multiplying its mass by the acceleration due to gravity (9.8 m/s^2). The torque due to the weight of the meter stick is given by the product of its weight and its perpendicular distance from the point of rotation, which is 100 cm.
Torque1 = (0.200 kg * 9.8 m/s^2) * 100 cm = 196 N * 1 m = 196 N·m
Next, let's calculate the torque due to the attached mass of 700 grams at the 40 cm mark. The torque due to this weight is given by the product of its weight and its perpendicular distance from the point of rotation, which is 60 cm.
Torque2 = (0.700 kg * 9.8 m/s^2) * 60 cm = 6.86 N * 0.6 m = 4.116 N·m
To maintain torque equilibrium, the sum of the torques acting on the meter stick must be zero:
Torque1 + Torque2 - F * 100 cm = 0
Substituting the values we found:
196 N·m + 4.116 N·m - F * 100 cm = 0
We can rearrange this equation to solve for F:
F * 100 cm = 196 N·m + 4.116 N·m
F * 100 cm = 200.116 N·m
F = 200.116 N·m / 100 cm
F = 2.00116 N
Therefore, the magnitude of force F is approximately 2.00116 Newtons.
To find the magnitude of force N, we can use the fact that the sum of all the forces acting vertically must equal zero since the meter stick is in equilibrium.
N - (Weight of the meter stick) - (Weight of the attached mass) = 0
The weight of the meter stick is equal to its mass multiplied by the acceleration due to gravity, which is 0.200 kg * 9.8 m/s^2 = 1.96 N.
The weight of the attached mass is equal to its mass multiplied by the acceleration due to gravity, which is 0.700 kg * 9.8 m/s^2 = 6.86 N.
N - 1.96 N - 6.86 N = 0
N = 1.96 N + 6.86 N
N = 8.82 N
Therefore, the magnitude of force N is approximately 8.82 Newtons.