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         To start, we'll identify all of the forces on the block, and on the knot that joins the three
         ropes, as shown here:




         The block has two forces on it. Its weight, mg straight down, and T the tension in the
                                                                                  3
         rope pulls straight up. We know that F=ma, so for this block we can write

                                                   T3-mg=ma



                                                                                                     Equation 1


                                                                                                          2
         T is positive and up, mg is negative and down. Since the block isn't moving, a=0 m/s ,
         so this equation becomes

                                                    T3-mg=0



                                                                                                     Equation 2


                                                                     2
         So, right away, we see that T =mg, or T =2 kg (9.8 m/s ), or T =19.6 N .
                                        3           3                        3


1 of 4                                                                                                 10/21/2012 8:15 PM
http://www.fearofphysics.com/Probs/mech037.html


         Now the knot is sort of a "transfer point" for the forces, so at the knot we can write


                                                      T1+T2+T3= 0




                                                                                                   Equation 3



         It's all equal to zero since the knot is stationary.

         Now, this is a vector equation, which you cannot deal with directly. Instead, we must
         look at the x and y components of each vector. Here are the components (we'll call up +,
         down -, left -, and right +):

         For T : T =-T cos          and T =T sin            .
              1    1x    1      1        1y       1    1

         For T : T =T cos           and T =T sin        .
              2    2x   2       2       2y    2        2

         For T : T =0 (since it points down only), and T =T .
              3    3x                                           3y   3

         Plugging in numbers, you'll have

         For T : T =-T cos50 and T =T sin50.
              1    1x    1              1y    1

         For T : T =T cos60 and T =T sin60
              2    2x   2               2y    2

         For T : T =0 (since it points down only), and T =19.6 N
              3    3x                                           3y

         Now, since we have all of the vector components, we can set up component forms of
         Equation 3 like this:

         x-direction:

                                                  -T1cos50+T2cos60=0



                                                                                                   Equation 4



         y-direction:



2 of 4                                                                                               10/21/2012 8:15 PM
http://www.fearofphysics.com/Probs/mech037.html



                                          T1sin50+T2sin60=19.6 N



                                                                                                Equation 5



         So if you look carefully, we have 2 equations and 2 unknowns, T and T .
                                                                          1         2

         Solve Equation 4 for T to get
                               1


                                                    T1=0.77T2



                                                                                                Equation 6



         Now, plug this value for T into Equation 5 to get
                                     1


                                         0.77T2sin50+T2sin60=19.6 N



                                                                                                Equation 7



         Solve for T to get:
                    2


                                                       19.6 N
                                               T2=
                                                  0.77sin50+sin60


                                                                                                Equation 8



         or

         T =13.46 N .
          2

         Then, from Equation 6 you can get T =10.36 N .
                                                1

         So, the three tension are

         T =10.36 N , T =13.46 N , and T =19.6 N .
          1             2                  3



3 of 4                                                                                            10/21/2012 8:15 PM
http://www.fearofphysics.com/Probs/mech037.html


         200 completely solved physics problems for as little as 10 cents per problem! Click here.




                                        (c)2001-2008 FearOfPhysics.com




4 of 4                                                                                                  10/21/2012 8:15 PM

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Mech037

  • 1. http://www.fearofphysics.com/Probs/mech037.html www.fearofphysics.com To start, we'll identify all of the forces on the block, and on the knot that joins the three ropes, as shown here: The block has two forces on it. Its weight, mg straight down, and T the tension in the 3 rope pulls straight up. We know that F=ma, so for this block we can write T3-mg=ma Equation 1 2 T is positive and up, mg is negative and down. Since the block isn't moving, a=0 m/s , so this equation becomes T3-mg=0 Equation 2 2 So, right away, we see that T =mg, or T =2 kg (9.8 m/s ), or T =19.6 N . 3 3 3 1 of 4 10/21/2012 8:15 PM
  • 2. http://www.fearofphysics.com/Probs/mech037.html Now the knot is sort of a "transfer point" for the forces, so at the knot we can write T1+T2+T3= 0 Equation 3 It's all equal to zero since the knot is stationary. Now, this is a vector equation, which you cannot deal with directly. Instead, we must look at the x and y components of each vector. Here are the components (we'll call up +, down -, left -, and right +): For T : T =-T cos and T =T sin . 1 1x 1 1 1y 1 1 For T : T =T cos and T =T sin . 2 2x 2 2 2y 2 2 For T : T =0 (since it points down only), and T =T . 3 3x 3y 3 Plugging in numbers, you'll have For T : T =-T cos50 and T =T sin50. 1 1x 1 1y 1 For T : T =T cos60 and T =T sin60 2 2x 2 2y 2 For T : T =0 (since it points down only), and T =19.6 N 3 3x 3y Now, since we have all of the vector components, we can set up component forms of Equation 3 like this: x-direction: -T1cos50+T2cos60=0 Equation 4 y-direction: 2 of 4 10/21/2012 8:15 PM
  • 3. http://www.fearofphysics.com/Probs/mech037.html T1sin50+T2sin60=19.6 N Equation 5 So if you look carefully, we have 2 equations and 2 unknowns, T and T . 1 2 Solve Equation 4 for T to get 1 T1=0.77T2 Equation 6 Now, plug this value for T into Equation 5 to get 1 0.77T2sin50+T2sin60=19.6 N Equation 7 Solve for T to get: 2 19.6 N T2= 0.77sin50+sin60 Equation 8 or T =13.46 N . 2 Then, from Equation 6 you can get T =10.36 N . 1 So, the three tension are T =10.36 N , T =13.46 N , and T =19.6 N . 1 2 3 3 of 4 10/21/2012 8:15 PM
  • 4. http://www.fearofphysics.com/Probs/mech037.html 200 completely solved physics problems for as little as 10 cents per problem! Click here. (c)2001-2008 FearOfPhysics.com 4 of 4 10/21/2012 8:15 PM