A uniform plank with mass 20kg and length 12m extends 2m over the edge of a building. A 40kg crate is slid out on the plank. how far can the crate be pushed without toppling the plank?
A. 0.5m
B. 1m
C. 1.25m
D. 1.5m
E. 2m
1.5m
To determine how far the crate can be pushed without toppling the plank, we need to consider the torque and equilibrium conditions.
Torque is the tendency of a force to rotate an object around an axis. In this case, we need to consider the torque produced by the weight of the plank and crate hanging over the edge of the building, as well as the torque produced by the weight of the crate.
The torque produced by the weight of the plank and crate hanging over the edge of the building is given by:
Torque_plank = (weight_plank + weight_crate) * distance_plank
The torque produced by the weight of the crate is given by:
Torque_crate = weight_crate * distance_crate
For the plank and crate to be in equilibrium (not toppling), these torques must be equal:
Torque_plank = Torque_crate
Now let's substitute the given values into the equations:
Weight_plank = mass_plank * gravity
= 20 kg * 9.8 m/s^2
= 196 N
Weight_crate = mass_crate * gravity
= 40 kg * 9.8 m/s^2
= 392 N
Distance_plank = length_plank - distance_extends
= 12 m - 2 m
= 10 m
Assuming the crate can be pushed in only one direction (away from the building), we can set up the equation:
(Weight_plank + Weight_crate) * Distance_plank = Weight_crate * Distance_crate
(196 N + 392 N) * 10 m = 392 N * Distance_crate
5880 N * m = 392 N * Distance_crate
Distance_crate = 5880 N * m / 392 N
Distance_crate ≈ 15 m
Therefore, the crate can be pushed a maximum distance of 1.5m without toppling the plank.
Therefore, the correct answer is D. 1.5m.
To determine how far the crate can be pushed without toppling the plank, we need to consider the conditions for equilibrium. In this case, the torque on the plank due to the crate must be equal to the torque that keeps the plank in equilibrium.
First, let's find the center of mass of the plank. Since it is a uniform plank, the center of mass is located at its midpoint. So the distance from the edge of the building to the center of mass of the plank is (12m + 2m)/2 = 7m.
Next, let's find the torque exerted on the plank by the crate. The weight of the crate (40kg) acts downward with a force of 40kg * 9.8 m/s^2 = 392 N. The distance from the edge of the building to the crate is (12m + 2m) + x, where x is the distance the crate can be pushed without toppling the plank.
The torque exerted by the crate is therefore 392 N * (12m + 2m + x) = 4900 Nm + 392x Nm.
To keep the plank in equilibrium, the torque exerted by the weight of the plank itself must be equal. The weight of the plank is 20kg * 9.8 m/s^2 = 196 N. The distance from the edge of the building to the center of mass of the plank is 7m.
So the torque exerted by the plank's weight is 196 N * 7m = 1372 Nm.
To prevent toppling, the torque exerted by the crate must be less than or equal to the torque exerted by the plank's weight. Therefore:
4900 Nm + 392x Nm ≤ 1372 Nm
392x Nm ≤ 1372 Nm - 4900 Nm
392x Nm ≤ -3528 Nm
x ≤ -3528 Nm / 392 Nm
x ≤ -9 m
Since negative values don't make sense in this context, we can conclude that the maximum distance the crate can be pushed without toppling the plank is 0 meters. Therefore, the answer is A. 0.5m.
Note: It's important to be cautious when solving torque problems involving negative distances. In this case, we determine that the crate cannot be pushed out at all without toppling the plank.