A pole AB of length 10.0m and weight 800N from the end A and lies on horizontal ground. The end B is to be lifted by a vertical force applied at B calculate the least force that is required to do this.
Balancing the pole at A with tbe weight acting at cebtre of gravity 800(4) equals Force at B times 10
Then force at B equals 3200 divided by 10 equals 320N
Correct
To calculate the least force required to lift the end B of the pole, we need to consider the equilibrium of moments about point A.
Let's assume that the force applied at point B is F and the distance from point A to B is x.
The weight of the pole, acting vertically downward at its center of mass, can be considered a single force acting at its midpoint. The weight (W) can be calculated using the given weight of 800 N.
Given:
Length of the pole (AB) = 10.0 m
Weight of the pole (W) = 800 N
Since the pole is lying on horizontal ground, we can assume that there is no acceleration in the vertical direction. Therefore, the sum of the vertical forces must be zero.
Sum of vertical forces:
F - W = 0
Therefore, F = W
Now, let's consider the equilibrium of moments about point A. The moments on either side of point A must balance.
Sum of moments about point A:
(F * x) - (W * (10 - x)) = 0
Substituting the value of F from above:
(W * x) - (W * (10 - x)) = 0
Simplifying the equation:
Wx - W(10 - x) = 0
Expanding the equation:
Wx - 10W + Wx = 0
Combining like terms:
2Wx - 10W = 0
2Wx = 10W
Dividing both sides by 2W:
x = 10/2 = 5
By substituting the value of x back into the equation F = W, we can calculate the least force required to lift the end B.
F = W
F = 800 N
Therefore, the least force required to lift the end B is 800 N.
To calculate the least force required to lift the end B of the pole, we need to find the minimum force that will create enough torque to balance the weight of the pole.
Torque is the product of force and the perpendicular distance from the point of rotation (in this case, point A, as the pole is lying on the ground) to the line of action of the force. In this scenario, the torque created by the weight of the pole (800N) will be balanced by the torque created by the force applied at point B.
1. First, let's calculate the torque created by the weight of the pole. The weight is acting at the center of mass of the pole, which is at the midpoint of AB, or 5.0m from point A. Therefore, the torque created by the weight is:
Torque_w = weight * perpendicular distance
= 800N * 5.0m
= 4000 N⋅m
2. Now, let's calculate the torque required to balance the weight. As the weight is acting vertically downward, the force applied at B should act vertically upward to counteract the torque. Thus, the perpendicular distance from the force to point A is the length of the pole itself, 10.0m.
Torque_B = force_applied * perpendicular distance
= force_applied * 10.0m
3. Since these torques must balance each other, we can set them equal to each other:
Torque_w = Torque_B
4000 N⋅m = force_applied * 10.0m
4. Solve for the force_applied:
force_applied = 4000 N⋅m / 10.0m
force_applied = 400 N
Therefore, the least force required to lift the end B of the pole is 400 Newtons.