The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.
To find the answer, we need to know about the tension.
<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>
- Let's draw the free body diagram of the system using the given data.
- From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
- For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

- We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

- To find Ny, we need to find the tension T.
- For this, we can equate the net horizontal force.

- Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,

- Thus, the magnitude of the force that the beam exerts on the hi.nge will be,

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.
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Answer:
The last one, nicotine from tobacco
It's not so much a "contradiction" as an approximation. Newton's law of gravitation is an inverse square law whose range is large. It keeps people on the ground, and it keeps satellites in orbit and that's some thousands of km. The force on someone on the ground - their weight - is probably a lot larger than the centripetal force keeping a satellite in orbit (though I've not actually done a calculation to totally verify this). The distance a falling body - a coin, say - travels is very small, and over such a small distance gravity is assumed/approximated to be constant.