A uniform rod is 4m long and weighs 10 kg.if it is supported on a knife edge at one metre from the end what weight placed at that end keeps the rod horizontal?

A uniform rod weighs 10kg and with a load 5kg attached to one end,it balances on a knife edge at 2m from that end.the length of the rod is?

W=mg

5x2=[L/2-2] 10x10
L/2=1+2
L=6m

To determine the weight required to keep the rod horizontal when supported on a knife edge, we can use the principle of moments.

1. Start by drawing a diagram of the situation. The uniform rod is 4m long and supported on a knife edge at a distance of 1m from one end. Let's label the end where the weight will be placed as point A and the knife edge as point B.

A ------[]------------------ B
1m 3m

2. Calculate the weight of the rod. Given that the rod weighs 10 kg, we can find the force acting at its center of mass using the equation: weight = mass x gravity.

weight of rod = 10 kg x 9.8 m/s² = 98 N

3. Calculate the moment acting clockwise around point B due to the weight of the rod.

moment = weight of rod x distance AB

moment = 98 N x 3m = 294 Nm

4. Since the rod is in equilibrium, the clockwise moment due to the weight of the rod must be balanced by a counterclockwise moment due to the weight placed at point A.

5. Let's consider the weight to be placed at point A as weight W. The distance from point A to point B is 1m.

counterclockwise moment = weight W x distance AB

counterclockwise moment = W x 1m = W Nm

6. Equating the clockwise moment and the counterclockwise moment, we get:

294 Nm = W Nm

7. Solve for W to find the weight required to keep the rod horizontal:

W = 294 N

Therefore, a weight of 294 N placed at the end of the rod will keep it horizontal when supported on the knife edge.

To find the weight needed to keep the rod horizontal, we need to consider the balance of torques. The torque is the force applied to an object multiplied by the distance from the pivot point.

First, let's analyze the forces acting on the rod. We have the weight acting at the center of the rod, and we need to find the weight to be placed at the end of the rod.

Let W be the weight to be placed at the end of the rod (in Newtons).
The weight of the rod, which acts at its center, is 10 kg multiplied by the acceleration due to gravity (9.8 m/s²), so W_rod = 10 kg × 9.8 m/s² = 98 N.

The distance of the center of the rod from the pivot point is 4 m × 0.5 = 2 m. Since the weight of the rod acts at the center, we don't need to consider its torque for balance.

Now, let's calculate the torque due to the weight at the end of the rod. The distance from the pivot point is 1 m (as given in the question). So, the torque due to this weight is W × 1 m.

For the rod to be horizontal, the total torque must be zero. Therefore, we can set up the equation:

W × 1 m = 98 N × 2 m

Simplifying the equation, we find:

W = (98 N × 2 m) / 1 m
W = 196 N

Therefore, a weight of 196 Newtons needs to be placed at the end of the rod to keep it horizontal.