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A man supports himself and the uniform horizontal beam pulling the rope with a force T.The weights of men and the beam are 883 N and 245 N respectively.Calculate the tension T in the rope and the forces exerted by the pin at A.

Answer :

Answer:

T=502.5N

Ax=171.8N

Explanation:

The computation of the tension T in the rope and the forces exerted by the pin at A is shown below:

vertical forces sum = Ay + Tsin20 + T - 245 - 883 = 0

Now  

horizontal forces sum = Ax - Tcos70

Now Moment about B

-Ay × 4.8 + 245 × 2.4 + 883 × 1.8=0

Ay=453.6N

Now substitute in sum of vertical forces T=502.5N

Ax=171.8N

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Lanuelgo

a. The tension (T) in the rope is equal to 502.51 Newton.

b. The forces exerted by the pin at A is equal to 171.86 Newton.

Given the following data:

  • Weight of men = 883 N
  • Weight of beam = 245 N

To calculate the tension (T) in the rope and the forces exerted by the pin at A:

First of all, we would determine the vertical force by taking moment about point B as shown in the diagram.

[tex]-A_y \times 4.8 + 883 \times 1.8 + 245 \times 2.4 =0\\\\-4.8A_y + 1589.4 + 588 =0\\\\4.8A_y= 3237\\\\A_y=\frac{2177.4}{4.8} \\\\A_y= 453.63 \;Netwon[/tex]

The tension (T) in the rope would be calculated by the sum of the vertical component of forces, which is given by:

[tex]\sum F_x = A_y + Tsin20 + T - 245 - 883 = 0\\\\453.63 + 0.3420T + T -1128=0\\\\1.3420T = 1128-453.63\\\\1.3420T =674.37\\\\T =\frac{674.37}{1.3420}[/tex]

Tension, T = 502.51 Newton.

To find the forces exerted by the pin at A, we sum the vertical component of forces, which is given by:

[tex]\sum F_y = A_x - Tcos70 =0\\\\A_x =Tcos70\\\\A_x = 502.51 \times cos70\\\\A_x = 502.51 \times 0.3420\\\\A_x = 171.86\;Newton[/tex]

Read more: https://brainly.com/question/17742679

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