A uniform 7.9-m plank weighing 130 N lies on a platform with 3.5 m jutting off the platform. How far out on the plank from the edge of the platform can a 31-N dog walk without tipping the plank?
7.9/2 = 3.95
so center of plank is 3.95 from outer end
which is 3.95 - 3.5 = .45 inside the edge of the platform
so for equilibrium
130 (.45) = 31 x
x = 1.89 out from edge
To determine how far out on the plank from the edge of the platform the dog can walk without tipping the plank, we need to find the point at which the total torque is balanced.
1. First, let's find the torque produced by the weight of the plank. The weight of the plank can be calculated using the formula W = m * g, where W is the weight, m is the mass, and g is the acceleration due to gravity. The weight of the plank is given as 130 N.
2. The torque produced by the weight of the plank can be calculated as torque = force * distance, where torque is the rotational force, force is the weight, and distance is the distance from the pivot point (fulcrum). In this case, the fulcrum is at the edge of the platform. The distance is given as 3.5 m.
So, torque_plank = weight_plank * distance = 130 N * 3.5 m
3. Now, let's calculate the torque produced by the dog. The torque produced by the dog is given by torque_dog = force_dog * distance_dog, where force_dog is the weight of the dog and distance_dog is the distance of the dog from the edge of the platform.
We know the force_dog is 31 N. Let's assume the dog is x meters from the edge of the platform.
So, torque_dog = force_dog * distance_dog = 31 N * x
4. For the plank to remain balanced, the total torque produced by the weight of the plank must be equal to the total torque produced by the dog. So, we set torque_plank = torque_dog and solve for x:
130 N * 3.5 m = 31 N * x
455 N·m = 31 N * x
x = 455 N·m / 31 N
5. Simplifying the equation, x ≈ 14.68 m.
Therefore, the dog can safely walk out on the plank up to approximately 14.68 meters from the edge of the platform without tipping the plank.