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krw1

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**urgent help needed...I tried my best!~**

I tried my best for hours and need help help?

thanks for any help you can now?

A uniform, 31.9 kg beam of length 3.00 m is hanging horizontally as shown in the diagram below. It is pinned to a vertical wall at point B and supported by a uniform 21.4 kg diagonal brace that is pinned at point C on the beam and at point A on the wall. [Use g = 9.81 m/s2.]

http://lewis.chem.sfu.ca/res/tccfl/jac/24_Statics/Graphics/beam-2.png

There is a load W, with a weight of 157 N, hanging at the end of the horizontal beam. The distance BC is 2.13 m, and the distance AB is 1.68 m. (The distance BD is 3.00 m.)

What is the vertical component of the force on the horizontal beam at joint C? (Use "+" if the force component is up, "-" if it is down.)

Cy = 441.5 N

Hint for doing the next part:

Unlike the situation with a massless support, the "tension" in the diagonal brace will not be at the same angle as the brace. The second step in doing this problem is to use the free-body diagram for the brace, along with the 3rd-law reaction force to Cy that acts downward on the brace, to solve for Cx.

What is the horizontal component of the force on the diagonal brace at joint C? (Use "+" if the force component is right, "-" if it is left.)

Cx = You must be careful to have the correct direction for every force that acts on the brace, and the correct lever arm for each of them, in your free-body diagram for the brace. Do you have Cy pointing down, opposite to its direction when it acts on the beam?

Find the horizontal and vertical components of the force on the horizontal beam at the joint B. Use "+" for forces up or to the right, "-" for forces down or to the left.

Bx =

By =

Find the horizontal and vertical components of the force on the diagonal brace at the joint A. Use "+" for forces up or to the right, "-" for forces down or to the left.

Ax =

Ay =

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