To solve this problem, we need to apply the principle of moments. The principle of moments states that for a body to be in equilibrium, the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point.

Step 1: Identify the forces acting on the rod.
There are three forces acting on the rod:
- The weight of the rod, which acts at the center of the rod (50 cm mark) and has a magnitude of 6g Newtons (where g is the acceleration due to gravity).
- The weight of the 6 kg mass, which acts at the 40 cm mark and has a magnitude of 6g Newtons.
- The reactions at the supports, which we'll call R1 and R2. R1 acts at the 20 cm mark and R2 acts at the 80 cm mark.

Step 2: Choose a point to calculate moments about.
It's usually easiest to choose one of the supports, because then we don't need to consider the reaction at that support (since it has no moment about that point). Let's choose the left support (at the 20 cm mark).

Step 3: Calculate the moments about the chosen point.
- The weight of the rod gives a clockwise moment of 6g * 30 = 180g Newton-cm.
- The weight of the 6 kg mass gives a clockwise moment of 6g * 20 = 120g Newton-cm.
- R2 gives an anticlockwise moment of R2 * 60.

Step 4: Set the sum of the clockwise moments equal to the sum of the anticlockwise moments and solve for R2.
180g + 120g = R2 * 60
300g = R2 * 60
R2 = 300g / 60 = 5g Newtons

Step 5: Use the fact that the sum of the upward forces (the reactions) must equal the sum of the downward forces (the weights) to solve for R1.
R1 + R2 = 6g + 6g
R1 + 5g = 12g
R1 = 12g - 5g = 7g Newtons

So, the reactions at the supports are 7g Newtons at the left support and 5g Newtons at the right support.

Question

To solve this problem, we need to apply the principle of moments. The principle of moments states that for a body to be in equilibrium, the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point.

Step 1: Identify the forces acting on the rod. 
There are three forces acting on the rod: 
- The weight of the rod, which acts at the center of the rod (50 cm mark) and has a magnitude of 6g Newtons (where g is the acceleration due to gravity).
- The weight of the 6 kg mass, which acts at the 40 cm mark and has a magnitude of 6g Newtons.
- The reactions at the supports, which we'll call R1 and R2. R1 acts at the 20 cm mark and R2 acts at the 80 cm mark.

Step 2: Choose a point to calculate moments about. 
It's usually easiest to choose one of the supports, because then we don't need to consider the reaction at that support (since it has no moment about that point). Let's choose the left support (at the 20 cm mark).

Step 3: Calculate the moments about the chosen point.
- The weight of the rod gives a clockwise moment of 6g * 30 = 180g Newton-cm.
- The weight of the 6 kg mass gives a clockwise moment of 6g * 20 = 120g Newton-cm.
- R2 gives an anticlockwise moment of R2 * 60.

Step 4: Set the sum of the clockwise moments equal to the sum of the anticlockwise moments and solve for R2.
180g + 120g = R2 * 60
300g = R2 * 60
R2 = 300g / 60 = 5g Newtons

Step 5: Use the fact that the sum of the upward forces (the reactions) must equal the sum of the downward forces (the weights) to solve for R1.
R1 + R2 = 6g + 6g
R1 + 5g = 12g
R1 = 12g - 5g = 7g Newtons

So, the reactions at the supports are 7g Newtons at the left support and 5g Newtons at the right support.

Knowee AI · Accepted Answer

To solve this problem, we need to apply the principle of moments. The principle of moments states that for a body to be in equilibrium, the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point.

Step 1: Identify the forces acting on the rod. 
There are three forces acting on the rod: 
- The weight of the rod, which acts at the center of the rod (50 cm mark) and has a magnitude of 6g Newtons (where g is the acceleration due to gravity).
- The weight of the 6 kg mass, which acts at the 40 cm mark and has a magnitude of 6g Newtons.
- The reactions at the supports, which we'll call R1 and R2. R1 acts at the 20 cm mark and R2 acts at the 80 cm mark.

Step 2: Choose a point to calculate moments about. 
It's usually easiest to choose one of the supports, because then we don't need to consider the reaction at that support (since it has no moment about that point). Let's choose the left support (at the 20 cm mark).

Step 3: Calculate the moments about the chosen point.
- The weight of the rod gives a clockwise moment of 6g * 30 = 180g Newton-cm.
- The weight of the 6 kg mass gives a clockwise moment of 6g * 20 = 120g Newton-cm.
- R2 gives an anticlockwise moment of R2 * 60.

Step 4: Set the sum of the clockwise moments equal to the sum of the anticlockwise moments and solve for R2.
180g + 120g = R2 * 60
300g = R2 * 60
R2 = 300g / 60 = 5g Newtons

Step 5: Use the fact that the sum of the upward forces (the reactions) must equal the sum of the downward forces (the weights) to solve for R1.
R1 + R2 = 6g + 6g
R1 + 5g = 12g
R1 = 12g - 5g = 7g Newtons

So, the reactions at the supports are 7g Newtons at the left support and 5g Newtons at the right support.

A uniform 1𝑚 rod of mass 6𝑘𝑔 is being supported horizontally on two knife edgesplaced 20𝑐𝑚 from each end. If a mass of 6𝑘𝑔 is suspended at the 40𝑐𝑚 mark of therod, calculate the reactions at the supports

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