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# A crane suspends a 500-lb steel beam horizontally by support cables (with negligible weight) attached from a hook to each end of the beam. The support cables each make an angle of $60^\circ$ with the beam. Find the tension vector in each support cable and the magnitude of each tension.

## $\bullet$ The magnitude: 288.68 lb-weight$\bullet$ The tension vector form:$$\begin{array}{l}{\vec{T}_{1}=-288.68 g \cos \left(60^{\circ}\right) \hat{i}+288.68 g \sin \left(60^{\circ}\right) \hat{j}} \\ {\vec{T}_{2}=288.68 g \cos \left(60^{\circ}\right) \hat{i}+288.68 g \sin \left(60^{\circ}\right) \hat{j}}\end{array}$$

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stop about this question. Were given that the train suspends £500 steel being horizontally by support cables. And the cables doesn't have any weight attached from a hook or to each end of the b The support cables make an angle of 60° with the beam. And we need to find attention vector in each of the support beam and the magnitude. So since this is a symmetric figures, attention is gonna be same. So let the dinners the tensions. We're gonna make the components of the tension. We're here horizontal, vertical and the horizontal components of the vertical company is gonna be T signed 60 and the horizontal company is gonna be T cost 60 degree. And likewise for this one and the way it will be exactly downwards over here. So since this is an equilibrium, we're gonna say that uh over on the top the the vertical, some of all the vertical forces towards the Y axis is going to be zero. So that's gonna be T. Sign 60 plus T. Sign 60 is going to be equal to w. It means that to T signed 60 is route 3/2. That's going to be equal to W. Which means that T. Is equal to W over Route three and W is given us 500. So that's going to be 500 over Route three. That is the required value of 40. So this is the required magnitude which we needed to find. But if you want to find the tension in the factor form, then we're gonna write in terms of the competent. So that's gonna be T. Signed 60 I. 60 I plus T. Cause 60. Uh So that's gonna be uh detention factor in the vector form definitely. They're going to be equal. So let's call uh let's call this one left. So the left one is going to look like this. So he signs 60 is gonna be a 500 over route three times through three over those. 500 over three times rule three over to I Plus 500 over route three times 1/2 J. So that's going to be equal to 50 I. Cap Plus 2 50 over Route three Jacob. So this is the tension factor for the left one and for the right one. The only thing is going to change is the ex the I've actor because that's going to point towards the negative X. Axis. So the only difference is this sign is gonna, sorry this sign is going to change to negative sign. So we are going to say that the right side, the right side attention factor is gonna be negative 2 50 I plus 2 50 over through three checks. This is a required answer. Thank you

NIT Bhopal

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