In figure a) a 0.500kg block is suspended from a 1.25m long string. The ends attached to the ceiling are
separated by a distance of 1.00m.
a) What angle do the ends of the string make with respect to the ceiling?
b) What is the tension in the string?
c) The first block is removed and replaced by two 0.250kg blocks that are attached in such a way that
the length of each of the string segments (1,2,3) is equal. What is the tension in each segment of string?
So, keeping the positive directions as indicated, we note that the block is
not moving, so the sum of the forces in each direction must be zero.
∑ Fy = 0 → mg(− ŷ) + 2T sinθ (+ ŷ) = 0 and using the length of the string as
a second condition.
L1 + L2 = 1.25m; L1 = L2 → 2L cosθ = 1 → 1.25 cosθ = 1
so θ ≈ 36.9 wrt the horizontal.
Once the angle is known, the (y) equation may be solved and the tension established.
m
(0.5kg)(9.8 2 )
s = T ≈ 4.08N
mg = 2T sin θ →
2 sin(39.6)
Changing the way the strings are arranged changes the tensions in the strings, but the same basic
1.25m
= 0.416m .
procedure applies. L =
3
The ends of the string are still separated by 1m, so if you call the
T1
T3
horizontal projection of the string as d then
2d + 0.416m = 1m;d = 0.292m and d is related to the angle made
T2
by the string. 0.416 cosθ 2 = 0.292m → θ ≈ 45.5
We also know that the sum of the forces in each direction has not
changed – we can use the zero acceleration condition again for
each block. mg(− ŷ) + T1 sinθ 2 (+ ŷ) = 0
m
0.25kg(9.8 2 )
s ≈ 3.43N , and this must be the same as
So, T1 =
sin(45.5)
T3 since it makes the same angle with the ceiling.
If we look at the horizontal forces acting on the blocks we see that T1 cos(45.5) = T2 → T2 ≈ 2.4 N