two thin strings support a 200g metre rule 20cm and 30cm from the ends.calculate the tension in each string.
assuming cg is at the 50cm mark.
Summing moments about the 50 cm mark.
clockwisetorque+counterclockwise torque=zero
String1*30cm-String2*20cm=0
String1/string2=2/3
but the sum of vertical tensions+weight=0
String1+string2=2kg*9.8N/kg
sting2*2/3 + sting2=19.6N
string2 tension= 19.6*3/5
string1tension=19.6*3/5*2/3=19.6*2/5 N
To calculate the tension in each string, we need to use the principle of moments. The principle of moments states that for an object to be in rotational equilibrium, the sum of the clockwise moments must be equal to the sum of the anticlockwise moments.
Let's call the tensions in the two strings T1 and T2. The distance of the 200g (0.2 kg) metre rule from each end is given as 20 cm (0.2 m) and 30 cm (0.3 m). The weight of the metre rule acts downwards from its center, which is the midpoint of the length of the metre rule.
Now, since the system is in equilibrium, the sum of the clockwise moments must equal the sum of the anticlockwise moments.
Clockwise moments:
Clockwise moment of T1 = T1 * (0.2 m)
Clockwise moment of T2 = T2 * (0.3 m)
Anticlockwise moments:
Anticlockwise moment of weight = Weight of metre rule * distance of weight from the pivot = (0.2 kg) * (9.8 m/s^2) * (0.25 m)
Since the metre rule is not moving, the sum of the clockwise moments must be equal to the sum of the anticlockwise moments:
T1 * (0.2 m) + T2 * (0.3 m) = (0.2 kg) * (9.8 m/s^2) * (0.25 m)
Simplifying the equation:
0.2T1 + 0.3T2 = 0.049 N
Now, we also know that the sum of the tensions in the two strings is equal to the weight of the metre rule:
T1 + T2 = Weight of the metre rule = (0.2 kg) * (9.8 m/s^2)
Therefore, we have a system of two equations:
0.2T1 + 0.3T2 = 0.049 N
T1 + T2 = 1.96 N
We can solve these equations simultaneously to find the values of T1 and T2. Substituting the value of T2 from the second equation into the first equation:
0.2T1 + 0.3(1.96 - T1) = 0.049
Simplifying the equation:
0.2T1 + 0.588 - 0.3T1 = 0.049
Collecting like terms:
-0.1T1 = -0.539
Dividing both sides by -0.1:
T1 = 5.39 N
Substituting the value of T1 into the second equation to find T2:
5.39 N + T2 = 1.96 N
T2 = 1.96 N - 5.39 N
T2 = -3.43 N
Note: The negative value of T2 means that the direction of tension in the second string is opposite to the assumed direction. Therefore, we take the magnitude of T2, which is 3.43 N.
So, the tension in the first string (T1) is 5.39 N, and the tension in the second string (T2) is 3.43 N.