a uniform metal rod of length 80cm and mass 3.2kg is supported horizontally by two vertical spring balances at C and D. Balance C is 20cm from one end while balance D is 30cm from the other end. find the reading on each balance.
64n
When pivot at C:
Then C.M = (30 x 32) Ncm
A.C.M = 50D
C.M = A.C.M => 960/50 = 50D/50
D = 19.2N
C + D = 32N
C = 32 – 19.2
= 12.8N
Reading on C = 12.8N
Reading on D = 19.2N
yeah
To find the reading on each balance, we need to consider the forces acting on the metal rod and apply the principles of equilibrium.
Let's break down the problem step by step:
1. Draw a diagram: Sketch the rod with the spring balances C and D attached at the given distances from the ends.
C--------20cm---------|------------------30cm--------------D
| |
|------------------------80cm--------------------------|
2. Identify the forces: The rod is in equilibrium, which means the sum of all horizontal forces and the sum of all vertical forces acting on it must be zero.
- The weight of the rod acts vertically downwards from its center.
- The forces exerted by the spring balances act vertically upward.
3. Calculate the weight of the rod: The weight of an object can be calculated by multiplying its mass by the acceleration due to gravity (g).
Weight = mass × acceleration due to gravity
Weight = 3.2kg × 9.8 m/s²
Weight = 31.36 N
4. Calculate the distances from the supports: Since balance C is 20cm from one end and balance D is 30cm from the other end, we can calculate the distance of each balance from the center of the rod.
Distance of C from the center = 80cm / 2 - 20cm = 40cm - 20cm = 20cm
Distance of D from the center = 80cm / 2 - 30cm = 40cm - 30cm = 10cm
5. Set up the equilibrium equations: We need to set up two equations—one for the vertical forces and one for the horizontal forces.
Vertical Forces: The sum of the vertical forces must be zero.
Reading on balance C + Reading on balance D = Weight of the rod
Horizontal Forces: The sum of the horizontal forces must be zero.
No horizontal forces are acting on the rod since it is supported by the spring balances.
6. Solve the equations: We can substitute the values we have obtained into the equations to find the unknowns.
Reading on balance C + Reading on balance D = 31.36 N (equation 1)
Distance of C × Reading on balance C = Distance of D × Reading on balance D (equation 2)
Substituting the values:
20cm × Reading on balance C = 10cm × Reading on balance D (equation 2)
Now we have two equations with two unknowns. We can solve these equations simultaneously to find the readings on each balance.
Rearrange equation 2 to solve for Reading on balance C:
Reading on balance C = (10cm / 20cm) × Reading on balance D
Reading on balance C = 0.5 × Reading on balance D (equation 3)
Substitute equation 3 into equation 1:
0.5 × Reading on balance D + Reading on balance D = 31.36 N
1.5 × Reading on balance D = 31.36 N
Reading on balance D = 31.36 N / 1.5
Reading on balance D ≈ 20.91 N
Substitute the value of Reading on balance D into equation 3 to find Reading on balance C:
Reading on balance C = 0.5 × 20.91 N
Reading on balance C ≈ 10.45 N
7. Answer: The reading on balance C is approximately 10.45 N, and the reading on balance D is approximately 20.91 N.