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A load of 250 kg is hung by a crane's cable. The load is pulled by a horizontal force such that the cable makes a 30 angle to the vertical plane. If the load is in the equilibrium, calculate the magnitude of the tension in the cable​

Sagot :

Answer:

1250[tex]\sqrt{3}[/tex]

Explanation:

Let's look at this in a simple manner, because it is.

The crane weights 250Kg. Okay.

Since it is hung, there is the acceleration of gravity being applied on it (10m/s²)

Since F = m * a

F = 250 * 10

F = 2500

Now we know that the downward Force is 2500N.

To find the force that is being applied on that 30° angle, we can multiply our 2500N by cos30°, which happens to be [tex]\frac{\sqrt{3}}{2}[/tex].

Therefore, the force pulling the box in the cable's direction is:

[tex]\frac{2500\sqrt{3} }{2} = 1250\sqrt{3}[/tex]

If you have any questions feel free to comment.

Have a great one and mark brainliest if it helped, please

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