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A 500N person stands 2.5m from a wall against which a horizontal beam is attached. The beam is 6m long and weighs 200N. A cable is attached to the free end of the beam and makes an angle of 45 to the horizontal and is attached to the wall.
1)Determine the magnitude of the tension of the cable.
2)Determine the reaction force that the wall exerts on the beam.

Respuesta :

PART A)

Here by force balance along Y direction

[tex]Tsin45  + F_y = 500 + 200[/tex]

Force balance along X direction

[tex]Tcos45 = F_x[/tex]

now by torque balance

[tex]Tsin45 (6) = 500 (2.5) + 200 (3)[/tex]

[tex]T(4.24) = 1250 + 600[/tex]

[tex]T = 436.3 N[/tex]

PART B)

now from above equations

[tex]Tsin45 + F_y = 700[/tex]

[tex]F_y = 391.5 N[/tex]

[tex]F_x = Tcos45 = 308.5 N[/tex]

now net reaction force of wall is given as

[tex]F = \sqrt{F_x^2 + F_y^2}[/tex]

[tex]F = 498.4 N[/tex]

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batolisis

The magnitude of the tension of the cable is : 436.3 N

The reaction force that the wall exerts on the beam is : 498.4 N

A) Calculating the magnitude of the tension of the cable

considering the balancing of force along the Y-axis

Tsin 45 + Fy = 500 + 200  ---- ( 1 )

consider force along the X-axis

Tcos 45 = Fx

Applying torque balance

equation( 1 ) becomes

T* sin 45 (6) = 500*(2.5) + 200 * 3

Therefore :

T = ( 1250 + 600 ) / 4.24

  = 436.3 N

B) Calculating the reaction force that wall exerts on beam

considering  equation ( 1 ) above

T sin 45 + Fy = 700

Fy = 700 - T sin 45

    = 391.5 N

Fx = T cos 45

    = 308.5 N

Hence the reaction force that the wall exerts on beam is :

F = √308.5² + 391.5²

  = 498.4 N

Hence we can conclude that The magnitude of the tension of the cable is : 436.3 N and The reaction force that the wall exerts on the beam is : 498.4 N

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