A cylindrical steel wire is connected to a 15 m uniform rod...

A cylindrical steel wire is connected to a 1.5 m uniform rod with mass 25 kg. The steel is 0.5 m length with a cross-sectional diameter of 4.00 mm and could bear a maximum tension about 100 N. A 40 kg load is intended to place along the rod at a distance d. The Young modulus of the steel is 2.00×1011 N/m2. Determine: a. Where the load should be located as to make the wire about to snap! b. Determine the length of the wire if the situation in (a) is attained! c. Determine the strain of the wire if the situation in (a) is attained!

Transcript

0:00 Hello everyone.

00:01 For our perusal i have already drawn the diagram in the right -hand side and we'll be using it as we proceed throughout the solution.

00:08 So first of all, mass of rod is given as, so this is basically m is given as 25 kg and length of rod, length of rod l is given as 1 .5 meter.

00:29 We also have length of string length of string that is 0 .5 meter and cross sectional diameter is equal to 4 .00 millimeters okay also we have the cross sectional area is equals to pi d squared divided by 4 so from here what we are getting d we already have as 4 so this is becoming 4 pi millimeter square okay now load that is capital m is given as 40 kages moving forward total equilibrium condition about about point a in triangle a b sorry a triangle acd so in triangle ac d let us have a look at the triangle first so i'm just having a rough diagram of the particular triangle we are trying to evaluate.

01:49 So this angle here, this is basically our 60 minus alpha.

01:57 This angle here is our alpha and this is our 120 degree.

02:03 This is 0 .5 meters and this is 1 .5 meters.

02:08 Now using the sign rule, what we have is we'll be having sine alpha by 0 .5 is equal to sine 60 minus alpha divided by 1 .5 so from here what we are getting we are getting sine alpha is equals to this will be what 1 by 3 right 1 by 3 sine 60 cos alpha minus cos 60 sine alpha so just broke it into the formula we know some of these values let's put those so from there we are getting this and here we are getting it to 7 by 2 sine alpha is equal to root 3 by 2 cos alpha right so what does this mean that tan alpha will be let's look at tan alpha is equal to root 3 by 7 okay so from here we are getting alpha is equal to 13 .9 degrees.

03:29 Okay.

03:31 Now now tau is equal to zero.

03:41 So t times cost 30 degree times l minus mg d sine alpha is equal to zero.

03:55 Now put all the values you know.

03:57 So this is coming out to be 100 times root 3 by 2.

04:02 Times 1 .5 minus okay so this will be minus 40 times 9 .8 times d times sign 13 .9 right is equal to 0 right is equal to 0 so from here what we are getting we are getting d is equal to 1 .38 meters so now in the second point we will be finding the elongation of the wire...

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