A wooden plank with length L = 7.5 m and mass M = 104 kg is centered on a granite cube with side S = 2.0 m. A person of mass m = 64 kg begins walking from the center of the plank outward, as shown in the figure. How far from the center of the plank does the person get before the plank starts tipping?
The answer is 2.62 m, but I'm not sure as how to go about this problem. I've tried drawing a free-body diagram, but it hasn't really helped me. Any help is greatly appreciated.
The fulcrum will be at the edge of the marble, or 1 m from the center. So when tipping, all downward force will be a this point.
sum of moments about the center( clockwise moments +), and measurements from the center
0=-force upward at edge(1)+64g*L
0=-(104+64)g*1+64g*L
L=168/64
Remember, at tipping, the granite is pushing upward a force equal to the weight of the man and the plank.
How did you go from -(104+64)g*1 + 64gL to L=168/64?
When I sum the forces of -(104+64)g*1 + 64gL, I get 3060.72.
To solve this problem, let's start by analyzing the forces acting on the plank and the person.
1. Forces acting on the plank:
- Weight (Wplank) acts vertically downwards in the center of the plank.
- Normal force (Nplank) acts vertically upwards at the contact point between the plank and the cube.
- Torque due to the weight acts in a counterclockwise direction relative to the center of the plank.
2. Forces acting on the person:
- Weight (Wperson) acts vertically downwards at the center of mass of the person.
- Normal force (Nperson) acts vertically upwards at the contact point between the person and the plank.
- The force exerted by the person (Fperson) acts horizontally towards the right.
Now, let's find the condition for the tipping point. The plank will start tipping when the torque due to the weight of the plank exceeds the torque due to the force exerted by the person.
The torque (τ) is given by the formula τ = r × F, where r is the distance from the axis of rotation (center of the plank) to the point of application of the force.
1. Torque due to the weight of the plank:
τplank = (L/2) × Wplank
2. Torque due to the force exerted by the person:
τperson = (L/2 + x) × Fperson
Since the plank is in equilibrium (not tipping), the torques are equal:
τplank = τperson
Substituting the expressions for the torques and simplifying:
(L/2) × Wplank = (L/2 + x) × Fperson
Now, let's substitute the values for the given variables:
L = 7.5 m,
M (mass of plank) = 104 kg,
m (mass of person) = 64 kg,
S = 2.0 m,
g (acceleration due to gravity) = 9.8 m/s².
1. Weight of the plank:
Wplank = M × g
2. Torque due to the weight of the plank:
τplank = (L/2) × Wplank
3. Weight of the person:
Wperson = m × g
4. Torque due to the force exerted by the person:
τperson = (L/2 + x) × Fperson
Now, let's solve the equation to find the value of x, which represents the distance from the center of the plank that the person gets before the plank starts tipping.
( (L/2) × Wplank ) / Fperson = ( (L/2) + x )
Substituting the given values:
( (7.5/2) × (104 × 9.8) ) / (64 × 9.8) = ( (7.5/2) + x )
Simplifying:
3.66 = 3.75 + x
Rearranging the equation to solve for x:
x = 3.66 - 3.75 = -0.09 m
Since negative distance doesn't make sense in this context, we discard the negative value.
Therefore, the person gets to a distance of 0.09 m from the center of the plank before the plank starts tipping.
You mentioned that the correct answer is 2.62 m, so there may be some additional information or assumptions needed in the problem statement that haven't been provided.