Required information three loads p1 3000 kn p2 500 kn and p3...

Required information Three loads $P_1 = 3.000$ kN, $P_2 = 5.00$ kN, and $P_3 = 6.50$ kN are suspended as shown from the cable $ABCDE$ where $d_C = 4$ m. Determine the components of the reaction at $E$. The horizontal component of reaction at $E$ is kN$\rightarrow$. The vertical component of reaction at $E$ is kN $\uparrow$. (Round the final answer to two decimal places.)

Transcript

00:02 Hi, so i've been given this question to solve and this it consists of a in the problem there is given a cable with concentrated loads at three points on the cable so the loads are concentrated not distributed they are present at specific points in the cable and as you can see here in the diagram at point b a 2 kiloton force is acting at point c a 4 kiloton force is acting and at point b a 6 kilometers in force is acting and this cable is suspended from from two points that are at the same horizontal level okay so it is given that if if this dc this distance between between the point c to the level of the beams this dc is equal to four meters and we have to determine in part a the components of the reaction at e so these are two points a and e and because the cable is being hanging from these two points there's there will be some reaction forces at these two points because the cable will pull itself and to stay in equilibrium there will be some reaction forces okay so if this equal to four meter we have to find those reaction forces at e so at this at this point e there will be two two forces one would be e x in the horizontal direction and one would be ey in the vertical direction we have to find this net force.

02:15 This the resultant of this e x and ey we have to find that.

02:21 The components of the reaction at e so we have to find e x and ey.

02:38 Okay so whenever such a problem is given we first take the full body diagram free body diagram of the entire cable and from from that diagram we will we we make the moment about one of the one of the hanging points either a or e we make the sum of the moments there equal to zero you can quit the moment about any of the two points either a or e so similarly like there are reactions at e so the reactions are at a also similarly the reactions at a are this ax and ay okay, so let's solve this problem.

03:45 Let's make a free body diagram of the entire cable yes, so this is the free boy diagram of the entire cable and we are making the moment about a equal to zero because the the cable is in a rotational equilibrium so so the net torque on it acting should be zero so so so at any about any point because the because the cable is not rotating about a or about e it is in an equilibrium even if it is it would have been rotating it would it would not rotate with an acceleration so so at so we make the sum of the moments at a equal to zero okay so so about this point we take the moments like we we we take the net torque about this should be equal to zero and this torque is nothing but the sum of the moments about a and and the sum of the moments means that whatever forces are acting at this point so whatever forces they're not at this point whatever forces are acting that are some distance away from this point from a that with the position vector of that force and cross product to the position vector of of any force with respect to this point a and the force the sum of this r rectal f vector because r r r r r cross fictor is the moment of that force about a there's some should be equal to zero okay so because a x and a y won't have a moment about a because the r is zero here so the only forces that will the forces that will have moment are the other forces this force this force this force this force and a x y okay so let's let's make that equation so for this two kilo newton force the position vector is the position vector is always taken perpendicular to that force so this is this this distance here is the r for this force at b.

06:59 So this r here is 5 for this force.

07:04 So 5 meters and r are always points away from the, from this point a...

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