A train travels north at a speed of 50 m/s.
1 answer:
The apparent velocity is B) 48 m/s north
Explanation:
Here we have a problem of relativity of velocities.
In fact, the train is travelling north at a speed of
where this velocity is measured with respect to the ground.
At the same time, a passenger on the train is walking towards the rear (so, south) at a velocity of
where this velocity is measured with respect to the train, which is in motion in the opposite direction.
Therefore, the apparent velocity of the passenger with respect to an observer standing on the ground is:
And the direction is north, since this number is positive.
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The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:
We can apply the first Newton's law in x and y-direction.
If we do a free body diagram of the system we will have:
x-direction
All the forces acting in this direction are:
(1)
Where:
- T(1) is the tension due to the rope 1
- T(2) is the tension due to the rope 2
Here we just conclude that T(1) = T(2)
y-direction
The forces in this direction are:
(2)
Here W is the weight of the steel beam.
We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.
Knowing that T(1) = T(2) and W = mg, we have:
T(1) must be equal to 5479 N, so we have:
Therefore, the maximum angle allowed is θ = 37.01°.
You can learn more about tension here:
I hope it helps you!