QUESTION 5 Members ACE and BCD are connected by a pin at C...

Transcript

00:01 This question we are given here that ece members that ace and bcd ace and bcd are connected by a pin at c and the link d.

00:19 Here we can write link d we're talking about here that for the loading zone and determine the x component let's determine the x component the force exerted of the force exerted at c and the member we are talking about here is which is the bcd distance x and the y so the distance x we are talking about here is the 166 .94 unit here is the millimeter and the y component which is equal we can make 62 .23 millimeter so this is the component now here in this case we want to find the value of the cx so for that case we can write first that summation of f i summation of f i which is equal to 0 you can write a y so a y which is equal to 480 newton so a y is 480 newton here in this case we can write summation of n a which is equal to 0 so here we can write minus 480 multiple we can write 100 plus we can write b multiplied by 160 b multiplied by 160 so after that we can write we just simply find and we can write b which is equal to 300 newton b here is the 300 newton now we can talk about the summation of f x which is equal to 0 so here in this case we can write that b plus a x b plus a x which is equal to we can write the e x which is equal to minus 300 newton moment at c so we can write summation of here mc so we can write summation of mc which is equal to 0 and here we can write the fde fde multiple we can write sine alpha multiple we can write 250 plus we can write 300 multiple we can write 60 plus we can write 480 plus we can write 480 480 and the 100 here we can write 480 we're talking about here and here is 100 now after that we can write here in this question we just simplify and we can write fde which is equal we can write minus 264 divided by we can write sine alpha which is equal we can write minus 264 divided over we can write the value we are talking about here is is sign alpha but we don't know how what is the value of alpha so let's find out the first value of alpha here so here in this case you can write alpha which is equal we can write 10 inverse of the value of here y divided by value of x which is a substitute of x and y so value of x here is 62 .23 value of y here so here we can write 10 inverse of 62 .23 divided by here in this case the other value here is the aguiliof x 166 .96 .94.

03:57 Now we just simplify and we can write answer is just simplify and we can write.

04:02 So here in this case we can write 20 .44 degree.

04:07 So this is the value of alpha here.

04:09 Now here in this case this is the value of alpha we just substitute of value into this formula.

04:14 So here we can write signed 20 .44 now again we just simplify and we can write answer this right 755 .95 so this is the value of force here here and the force is in the newton here you can make summation of fx here you can make summation of fx which is equal zero here we want to find value of cx so here we can write cx minus fde cx minus fde multiply we can write cos alpha plus we can write 300 which is equal to now again we just simplify and we can write fd fd f which is equal to here we can write fde here we can write fde mullivore cause alpha minus we can write 300 again we just simply find we can write cf which is equal to fde so what is the value of fde here so if fde is 750 .955 so here we can write 755 .95...

Question

QUESTION 5 Members ACE and BCD are connected by a pin at C and by the link DE. For the loading shown, determine the x-component of the force exerted at C on member BCD. . The distances x and y are 166.94 mm and 62.23 mm, respectively.

Question

QUESTION 5 Members ACE and BCD are connected by a pin at C and by the link DE. For the loading shown, determine the x-component of the force exerted at C on member BCD. . The distances x and y are 166.94 mm and 62.23 mm, respectively.

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