Kodo Millet PPT made by Ghanshyam bairwa college of Agriculture kumher bhara...
Mech037
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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
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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:
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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
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