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:
- 1.07 ft
Explanation:
V1 = (-5, 7, 2)
V2 = (3, 1, 2)
Projection of v1 along v2, we use the following formula
=\frac{\overrightarrow{V1}.\overrightarrow{V2}}{V2}
So, the dot product of V1 and V2 is = - 5 (3) + 7 (1) + 2 (2) = -15 + 7 + 4 = -4
The magnitude of vector V2 is given by
=
So, the projection of V1 along V2 = - 4 / 3.74 = - 1.07 ft
Thus, the projection of V1 along V2 is - 1.07 ft.
so we need to find the direction of v2
Answer:
B
Explanation:
That's the answer. Hope it helped!
Answer:
Explanation:
From the conservation of mechanical energy
Solve to velocity v2
Velocity is about direction traveled in comparison to speed which is just distance with out direction