a chain that can support 2000 N is suspended between two trees. a 200 kg motor hangs from the middle. the slack in the chain makes the half slant upward 20 degrees. will the chain break?
To determine whether the chain will break or not, we need to analyze the forces acting on it.
Given:
- Maximum load the chain can support: 2000 N
- Weight of the motor (load): 200 kg
First, we need to find the tension in the chain caused by the weight of the motor. The weight of an object can be calculated using the formula:
Weight (W) = mass (m) × acceleration due to gravity (g)
In this case, the weight of the motor is given by:
W = 200 kg × 9.8 m/s² (acceleration due to gravity) = 1960 N
Now, let's consider the forces acting on the chain when it is slanting upwards at an angle of 20 degrees:
1. Tension force in the chain (T): The tension force acts vertically upward and is responsible for supporting the weight of the motor.
2. Weight of the motor (W): The weight acts vertically downward.
3. Normal force (N): The normal force acts perpendicular to the chain and opposes the tension force.
The sum of the vertical components of the forces must be zero for the chain to remain in equilibrium.
T sin(θ) + T sin(θ) - W = 0
where θ is the angle of slanting (20 degrees).
Substituting the values, we have:
T sin(20°) + T sin(20°) - 1960 N = 0
Simplifying the equation gives:
2T sin(20°) = 1960 N
Now, we can calculate the tension force (T):
T = 1960 N / (2 sin(20°)) ≈ 1960 N / (2 × 0.342) ≈ 2863 N
The calculated tension force is 2863 N, which is greater than the maximum load the chain can support (2000 N). Therefore, the chain will break under this load and configuration.
In summary, based on the given information, the chain will break when carrying a 200 kg motor with a slant of 20 degrees.