To solve these tension word problems, you will mostly need Newton's second law of motion.
Problem # 1
a 100 kg bucket is being lifted by a rope. The bucket started at rest and after being lifted 4 m, it is moving at 4 m/s. If the acceleration is constant, what is the tension on the rope?
Solution
The diagram below is useful in order to visualize the problem.
Observations:
F_{T} is the force of tension and it is represented with a bigger vector. This force has to be bigger since it is making the bucket go up.The only missing quantity is a. To get a, we can use one of the constant acceleration equations
Since we know V_{0}, V, and d, we can use V^{2} = V_{0}^{2} + 2adProblem # 2:
A rope is attached to a vehicle weighing 500 kN. A very strong man puts the other end of the rope inside his mouth at an angle of 30 degrees from the horizontal. If the strong man is able to move the vehicle with a constant force 1 m away with a speed of 0.25 m/s, what is the tension on the rope. (Use g = 10 m/s^2)
Solution
This situation is shown below with graph.
Newton's second law gives F = ma
Since the vehicle is only moving along the x axis, the only force on the vehicle is along the x axis.
We get F_{x} = ma_{x}Since the man is moving the vehicle with a constant force, the acceleration is constant. Therefore, we can use the following equation again to get the acceleration.
V^{2} = V_{0}^{2} + 2adMar 16, 17 03:15 PM
Great lesson about the law of reflection. Crystal clear explanation
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