I apologize for not further labeling the diagram but cramster is not working wel

Question

I apologize for not further labeling the diagram but cramster is not working well for me right now and I am having technical difficulties. Anyways, the vertical force pointing down is Q. Q = 480N. the slanted force pulling to your right with the small triangle is force P. The small triangle represents a 3-4-5 triangle. The large triangle has a hypotenuse of cable AC. C is the point wich connects the cables and forces. The small leg of the large triangle is cable BC. BC is on the same line of action as force Q, straight down. The distance between point A and B is 600 mm (x) and the height of point A is 250 mm (y).

The problem is the following: Two cables are tied together at point C and loaded (Q = 480N). Determine the range of values of force P for which both cables remain taught.

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Explanation / Answer

let = tan-1(250/600), = tan-1(3/4)

Fx = Pcos - TCAcos = 0, so TCA = Pcos/cos 0, P 0

Fy = Psin + TCAsin + TCB - Q = 0,

TCB = Q - Psin - TCAsin = Q - Psin - (Pcos/cos)sin

= Q - Psin( + )/cos 0

so P Qcos/sin( + ) = 514 N

answer: 0 P 514 N

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