Hello students, welcome to engineers Academy. Now we are going to solve this problem which states that a container is supported by three cables that are attached to a ceiling as shown. Determine the weight of the container knowing that the tension in cable AB is 6 kilo Newton.
and the second problem says that determine the weight w of the container knowing that the tension in cable ad is 4.3 kilonewton so the method for both the problem will remain the same i will solve this when the tension in cable ab is 6 kilonewton you guys can use the same method for solving this particular problem so since this weight is in equilibrium so this problem is related to the equilibrium and the sum of all these three tensions the vector sum of all these three tensions must be equals to zero so we can say that the tension in a cable a b plus the tension in cable a c plus the tension in cable a d plus the weight this must be equals to 0 so in order to apply this equation the vector sum we need to find we need to represent all these tensions in their Cartesian vector representation form we need to write all these tensions and the weight in terms of I J and K components so we can write TAB Cartesian vector this will be equal to TAB magnitude times the unit vector from A to B lambda AB and we know that this is the magnitude TAB and lambda AB is the position vector from A to B divided by the magnitude of the position vector from A to B. so again we will find the position vector from a to b using that same method that we have learned in the previous videos we will start traveling from a along the x y and z to reach that point b so from a we need to travel 600 mm in the positive y 600 mm in the positive y so we will reach here and then from here we need to travel this distance and this distance is 450 mm and this is in the negative x so we will reach that point b this will be tab magnitude so we need to travel uh 600 mm in the neck in the positive y so i will write plus 600 in the positive j 600 in the positive j and then 450 mm in the negative x so minus 450 in the in the negative i and there is no need to travel in the z so plus 0 k 450 square plus 600 square plus 0 square this gives a 750 so this is 750 mm and there is no need to convert the units since since all these are in mm and this is in mm so mm will cancel out with mm and as we know that the unit vector is always unit less so we will the dimensions will cancel out and this will this is a unit less fraction so we can say that TAB will be equal to now we need to write this component divided by this this by this this by this and multiply by tab so we can write that tab Cartesian vector will be equal to minus 450 divided by 750 TAB and this is in the in the I plus 600 divided by 750 TAB in the J and 0 so the K component will become 0 so TAB into 0 divided by 750 will become 0 so 0 K so 450 450 divided by 750 this gives us 0.6 so we can say that this is minus 0.6 t a b i and 600 divided by 750 this gives us 0.8 so plus 0.8 t a b j plus 0 k so this is the cartesian vector representation of t a b or we can say this is t a b vector in terms of its i j and k components similarly we need to write t a c as a cartesian vector so t a c vector this is again equal to t a c magnitude times the unit vector from a to c lambda a c and this is t a c and lambda a c is the position vector from a to C divided by its magnitude so we can write the t ac cartesian vector this is equal to t ac magnitude again from a we need to travel 600 mm in the positive j and then 320 mm in the negative z and there is no need to travel in the x so we can write that this will be 0 i plus 600 j in the j in the positive y we need to travel 600 mm and in the z in the negative z we need to travel 320 mm so minus 320 k and we can find the magnitude by taking the square or the sum of the square of all the components and then taking the square root so this will be 0 square plus 600 square plus 320 square this gives us 680. so this is 680 mm this is 680. and now we can write t ac cartesian vector this is equal to so i component is 0 0 i plus 600 divided by 680 t ac this is this is not vector this is magnitude this is magnitude right so t ac j minus 320 divided by 680 t ac k and let's simplify this 600 divided by 680 gives us 0.882 so this is zero i 0.882 t a c j and 320 divided by 680 gives us 0.470 or we can say 0.471 T A C K. This is the Cartesian vector representation of the tension from A to C. Similarly from A to D. So T A D. this will be equal to Tad magnitude times lambda a to d and again Tad lambda a to d is the position vector from a to d divided by its magnitude so Tad magnitude and the position vector from a to d So now from A again we need to travel 600 mm in the positive J.
So plus 600 J. So we will reach here and then from here we need to travel this distance 500 mm. If I extend this line. So we need to travel 600 mm in the positive Y that is in the positive J then 500 mm in the positive X and then 360 mm in the positive Z. So in the X we need to travel this distance which is 500 mm.
So plus 500 I and then from here I need to travel 360 mm in the positive Z so plus 360 in the positive K and again we need to find the magnitude so that is 500 square plus 600 square plus 360 square under the square root it gives us 860 so 860 magnitude so tad vector is equal to 500 divided by 860 this gives us 0.581 this is a multiplication right 0.581 t a d i plus 600 divided by 860 0.698 j and 360 0.419 in the positive k so now since this weight is in equilibrium it is stationary so the this equation is valid the vector sum of all these tensions plus the weight will be equal to zero or we can say that the sum of all the x component must be equal to zero the sum of y components must be equal to zero and the sum of the z components must be equal to zero the weight is zero i minus w j so the weight is acting in the negative j and its k component is zero as well so this is the cartesian vector representation of the weight so now let's add all the x component the sum of all the x component is equals to zero or we can say the sum of all the i components is equals to zero so we need to add up all the i components so 0.60 ab or we can say minus 0.60 ab plus zero plus this 0.581 t ad this must be equals to zero so minus 0.60 ab plus zero plus zero point five eight one t a d this is equal to zero and the magnitude of the tension in cable a b is given this is given six kilo newton so now we know it's this is very important that t a b magnitude given which is 6 kilo Newton this is very important now we got the sum of all the I components is this and which TAB magnitude is known so if TAB magnitude is known we can find the magnitude of TAD so from this we can say that 0.581 TAD is equal to plus 0.6 Tav. Now Tav is 6 kilo Newton. So dividing both sides by 0.581, we will get Tad magnitude.
So now Tad is 0.6 multiply 6 divided by 0.581. This gives us Tad equals to 6.19 or we can say this is 6.20 kilonewton since this 6 is in kilonewton so we will get the answer in kilonewton so tad is 6.20 kilonewton similarly the sum of all the j components must be equals to 0 so we can say that the sum of all the y components is equals to 0 we need to add up all the j components so this you the J component of TAB we can say plus 0.8 TAB plus the J component of TAC which is 0.882 TAC plus this and we need to write uh tad with with these two components as well right so this is by mistake this is into tad and this is tad so plus 0.698 tad and plus this this is minus the weight so minus the weight this is equals to zero so now in this equation tab is known tad is known tac is not known and the weight is not known so in this equation there are two unknowns so before going for further solution this will not give us the solution since in this equation we have two unknowns so you We will go for the sum of all the components in the Z and then we will come back to this equation in order to solve this for the weight. So we can add up all the, we can say the sum of all the Z components must be equals to 0. Or we can say the sum of all the K components must be equals to 0. So again we can see that this is 0K minus 0.471 TAC and minus 0.419 TAD and this is 0. so the Z component of TAB is 0 the Z component of TAC is minus 0.471 TAC and the Z component of TAD is 0.419 TAD and the Z component of the weight is 0 as well so plus 0 this is equal to 0 Now we know TAD which is 6.2 we can put it here and we will get TAC which is which was unknown.
So this is minus 0.471 TAC equals to minus 0.419 TAD. TAD magnitude is 6.20 dividing both side by minus 0.471 we will get t ac so t ac is minus 0.419 into 6.20 divided by um minus 0.471 we get t ac equals to 5.52 kilo newton since 6.2 is in kilo newton so t ac is 5.52 kilo newton and similarly uh putting all these weight in this equation will give us the weight which is required if I bring this weight to the other side of equation it will become positive and all these thing will become will be equal to weight so we can write this equation like this this will be equal to the weight so 0.8 times TAB and this and so I will do all this in calculator so 0.8 into TAB TAB magnitude is 6 kilo Newton that is given in the problem statement plus 0.882 TAC is this 5.52 0.8 tab 0.882 tac and 0.698 tad so plus 0.698 tad is 6.2 so into 6.2 0 and the weight is approximately equal to 14 kilo newton so weight is equal to 14 kilo newton approximately so now if the tension in cable a b is 6 kilo newton and the system is in equilibrium then the weight that the whole system can support is 14 kilo newton and tad magnitude is 6.20 kilo newton and the tension in rope or cable ac is 5.52 kilo newton so this is the solution of this particular problem i hope it will help you in your learning uh you guys must solve this problem and let me know the answer in the comment section below answer for this particular problem if the tension in cable AD is 4.3 kN. Do subscribe to Engineers Academy for the solution of such more problems from Vector Mechanics for Engineers by Baron Johnston.