Two blocks of masses 2 kg and 3 kg are
connected by a metal wire going over a smoother pulley as
shown in figure. The breaking stress of the metal is
2 x 9°Nm-2. What would be the minimum radius of the wire
used if it is not to break? Take, g = 10ms-2.
Answers
Answer:
Explanation:T=
m
1
+m
2
2m
1
m
2
g
=
1+2
2×1×2
×10 N
=
3
40
N
If r is the minimum radius, then breaking stress =
πr
2
3
40
⇒
3πr
2
40
=
3π
40
×10
6
⇒r
2
=
10
6
1
or r=
10
3
1
m
⇒r=
10
3
1
×10
3
mm=1 mm
Answer:
question not clear , ill answer asuming breaking s = 2 x 10^9 and the weigths are hang verically coz no picture is available
6 x 10^-4 m
Explanation:
breaking stress = Force/ area
force is equal to tension, here 24 newton (i dont have the patience to explain that)
so from above eqn area is 12 x 10^-9
equate it to ( area = 3.14 x r x r)
so approx r = 6 x 10^-4
I used heavy approximation ,double check the calculation. hope this helps
PS the tension part is too complicated to tell here sry
but use. tension= (m1 x m2 x 2g)/ (m1+ m2)
only if both are hanging vertically and not at angle. if its at an angle sin of angles come in, also again I cant see image of question