Question

Two wires are tied to the 400g sphere shown in fig

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Answer #1
Concepts and reason

The concept used to solve this problem is based on tension in the string.

Initially, the angle made by the wire can be calculated by using the relation between sine of angle formula. Later, the radius can be calculated by equating the tension formula along vertical and horizontal axis. Finally, the tension in the upper and lower wire can be calculated by solving the tension equation.

Fundamentals

The expression for the angle between wire and the midpoint line is,

sin 0 =

Here, is the angle, is the distance from midpoint to one end of the wire, andis the length of the wire.

The expression for the tension along x-direction is,

(T. +T;)cos 0 =

Here, are the tensions in the upper wire and in the lower wire respectively, is the mass, is the velocity, andis the radius.

The expression for the radius is,

r=Ľ - x?

The expression for the tension along y-direction is,

T, sin 0 =T, sin + mg

Here, is the gravitational acceleration.

(a)

The diagram for the angle is depicted below:

The expression for the angle between wire and the midpoint line is,

sin 0 =

Substitute 0.5 m
for, 1 m
for

sino - (0.5 m)
(1 m)
= 30°

The expression for the radius is,

r=Ľ - x?

Substitute 1 m
for, 0.5 m
for.

r= (1 m) -(0.5 m)
= 0.866 m

The expression for the tension along x-direction is,

(T. +T;)cos 0 =

The expression for the tension along y-direction is,

T, sin 0 =T, sin + mg

Substitute 30°
for.

(T; +T, )cos(30) = mv
7, +1, - (my?)
r(13/2)
- 2(mv?)
13
…… (1)

T, sin (30)=T, sin (30)+mg
T.-T, = 2mg
…… (2)

Add equation (1) and (2), the expression for the tension in the upper wire is,

E14
Am + 8u = 1

Substitute 400 g
for, 8.4 m/s
for, 9.8 m/s
for, 0.866 m
for.

1x 10 kg
(8.4 m/s)
- () ).makeup
1g
(0.866 m) 3
= 22.74 N

(b)

The expression for the tension in the lower wire is,

T, -T, = 2mg

Substitute for, 22.74 N
for, for,

N 68 =
(su80)|(1,041)](800) a= -(vc Ze)

Ans: Part a

The tension in the upper wire is22.74 N
.

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