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Please help with tension question!

  1. Aug 21, 2005 #1
    Hey everyone,
    Please can someone help me with this question coz I am tryna help out a friend who can't do it but I am completely stumped!

    a) Draw an appropriate force diagram for the forces acting at B (only) and use the Sine Rule to show that the tension in the cable is 1.3014kN to 4 decimal places. Determine also the compressive force in the bar, to the same accuracy.

    b) Determine the minimum diameter of the cable if the maximum allowable tensile stress in the cable is 17 MN/m^2.

    Picture which goes with the question is attached, please can someone help me coz it's driving me crazy! Even telling me what formulae to use would be good coz I'm not a physics student so don't know where to start! Any hints would be much appreciated!

    Sarah xx :uhh:
     

    Attached Files:

  2. jcsd
  3. Aug 21, 2005 #2

    HallsofIvy

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    There is given a 1.5 kN downward force at B. The tension force in the cable and compression force in the bar must be such that the net vertical component is 1.5 kn upward and the net horizontal component is 0. Since the triangle- cable, bar, wall- has wall side vertical, you can imagine that being the force triangle as well. The "length" of the cable side is "X", the "length" of the bar side is "Y", and the "length" of the wall side is 1.5 kN. Since you know the angles, you can use the sine law to calculate X and Y, the tension in the cable and the compression in the bar.
     
  4. Aug 21, 2005 #3
    These Forces are in equilibrium and aren't moving so when you setup your (Sum)F = ma in the x and y direction the right side of the equation will go to 0. Leaving you with (Sum)Fx = 0 and (Sum)Fy = 0.
    From there I'm not to sure with a tensile factor what equation to use but I'm guessing that you can relate it towards that factor, tension, and diameter.
     
  5. Aug 21, 2005 #4
    Hi,
    I think I know how to do this.
    Have a look at the picture I've attached.
    First thing you have to realise is there are two forces supporting the 1500N- a tension force in the cable and a compression force force in the bar.
    The first thing you need to do is resolve the tension and compression forces into their vertical and horizontal components, Ty and Tx and Cy and Cx.
    This is where the sine rule bit comes in. I could have made my picture clearer- imagine the red line Ty slid over so that it is above the blue line Cy. So we have a red and blue triangle. Remember sin= opposite/ hypotenuse, so
    Ty= Tsin30, Tx= Tsin60.
    Similarly,
    Cy=C sin 37 and Cx= C sin53.
    Now since everything is at rest, all the forces must balance, so in the vertical (y) direction,
    Tsin30+Csin37=1500 [1]
    and in the horizontal (x) direction,
    Tsin60=Csin53 [2]
    so now you need to rearrange [2] to find an equation for C in terms of T and substitute into [1]
    this should let you find T. To find C, just put your value for T into [1]


    For the second part you know that the cable will support a certain force per area and you know the force it has to support (T), so you can work out what cross sectional area of cable you need and so what diameter of cable you need.

    hope that helps. :biggrin:

    apologies in advance if anyone else thinks I've given too much away.
     

    Attached Files:

  6. Aug 21, 2005 #5
    Thanx for the help, ive now shown the tension in the cable but am still confused about the compressive force! I forgot to write out that this was the intro to the question:

    A shop sign weighing 1.5kN is to be hung from point B on a support fixed to the shop wall AC, as shown in the figure Q3. AC is a rigid steel bar and AB is a steel cable.

    I am confused coz shouldnt BC be the steel bar to hold the cable in place? and then would i just work out the compressive force in the same way i worked out the tension of AB? It definitely says AC is the steel bar so I'm confused! Also any help for part b? Do i use the ultimate tensile stress equation and if so is maximum load just 1500kg? I wish i understood this physics stuff!!
     
  7. Aug 21, 2005 #6
    Just gone through it based on BC being the steel bar and came out with 1.4112 kN being the compressive force. On part b got 11mm (to the nearest mm) being the minimum diameter? Very unsure about the second one coz dunno if i was using the right equation but it sounds kinda right? Can anyone tell me if they r right? Thank u so much everyone for ur help, ur lifesavers!! My friend is gonna love me forever!! xx :-)
     
  8. Aug 21, 2005 #7
    hi
    your answer for the first part looks correct.
    I think there must be a typo in the original question. it says that AC is the wall AND the steel bar? I think it's safe to assume that it means BC is the steel bar.

    For the second part, I think you did it correctly: you used

    tensile stress=force/area.

    where area is the cross sectional area of the cable. ?

    we know that the maximum tensile stress can be 1.7x10^7 N/m^2 and we know it has a force (tension) acting on it of 1.3014x10^3 N (it's not 1500 N since the cable doesn't have to support the whole weight because the steel bar is taking some of the weight as well), so you can work out the cross sectional area of cable needed and from there the diameter.
    Try the maths again - you weren't far off-I think it works out at a little less than 11mm
    good luck :biggrin:
     
    Last edited: Aug 21, 2005
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