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.
Learn more about the tension here:
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The average adult in the us spends 24 hours watching televistion each week
Answer:
Ff = 19.6 N
Explanation:
So since its saying whats the minimum F to move the block, we will use static friction (0.5).
We will use the equation for force of friction, which is Ff = uFn
Ff = (0.5)(4)(9.8)
Ff = 19.6 N
this is the minumum force needed to move the block, as that is the frictional force. You would need to apply a minimum force of 19.6 N to move the block