A pole AB of length 10m and weight 600N has its centre of gravity 4m from the end A, and lies on horizontal fround.calculate the force required to begin to lift the end.why would this force applied at the end A,not be sufficient to lift the end A?
F * 10 = 600 * 4
F = 600 * .4 = 240 N
if you tried to lift at the other end
F2* 10 = 600 * 6
F2 = 360 N
Clock wise moment=600×4=2400NM Anticlockwise moment=p×10=p10NM Equilibrium:2400=10pNM divide both sides by 10 Therefore p=240
To calculate the force required to begin lifting the end of the pole, we need to consider the moment equilibrium.
The moment equilibrium equation is given by:
Sum of clockwise moments = Sum of anticlockwise moments
Let's assume that the force required to begin lifting the end is F, and the distance of F from end A is x (which is equal to the length of the pole, 10m).
Clockwise moments:
There is only one clockwise moment, which is the weight of the pole acting at its center of gravity (4m from end A). The moment can be calculated as:
Clockwise moment = weight x distance = 600N x 4m = 2400 Nm
Anticlockwise moments:
There are two anticlockwise moments. The first one is the force F required to begin lifting the end A, acting at a distance of x (10m) from end A. The moment can be calculated as:
First anticlockwise moment = force x distance = F x 10m = 10F Nm
The second anticlockwise moment is the weight of the pole acting at the end A. The moment can be calculated as:
Second anticlockwise moment = weight x distance = 600N x 10m = 6000 Nm
We can now set up the moment equilibrium equation:
Sum of clockwise moments = Sum of anticlockwise moments
2400 Nm = 10F Nm + 6000 Nm
Simplifying the equation:
-3600 Nm = 10F Nm
Dividing both sides by 10 Nm:
-360 N = F
Therefore, the force required to begin lifting the end of the pole is -360 N (negative sign indicating the force is acting in the opposite direction).
Now, why would this force applied at the end A not be sufficient to lift the end A?
The force required to lift the end A is calculated to be -360 N. However, due to the weight of the pole acting at the end A (600N) and its distance from the pivot point (10m), the resulting anticlockwise moment is greater than the clockwise moment. This means that the force applied at the end A is not enough to overcome the weight of the pole and lift the end A. Additional force is needed to generate a sufficient anticlockwise moment and initiate the lifting process.
To calculate the force required to begin to lift the end of the pole, we need to understand the concept of torque. Torque is the measure of the force's ability to rotate an object around an axis, and it is given by the product of the force and the distance from the force to the axis of rotation.
In this case, the axis of rotation is the center of gravity of the pole, which is located 4m from the end A. To lift the end A of the pole, we need to apply a force at the end that creates a torque greater than the torque caused by the weight of the pole.
The torque caused by the weight of the pole can be calculated by multiplying the weight (600N) by the distance from the center of gravity to the end A (4m). So, the torque due to the weight of the pole is:
Torque = Weight × Distance = 600N × 4m = 2400Nm
To lift the end A, we need to apply a force with a greater torque than 2400Nm. However, if we apply the force at the end A, the distance from the force to the center of gravity is 0m. Therefore, the torque produced by this force is:
Torque = Force × Distance = Force × 0m = 0Nm
As we can see, the torque created by the force applied at the end A is zero. Since the torque required to lift the pole is 2400Nm, the force applied at the end A will not be sufficient to lift it.
To calculate the force required to begin lifting the end A, we can rearrange the torque formula and solve for force:
Force = Torque / Distance
In this case, the torque required to lift the pole is 2400Nm, and the distance from the force (applied at the end A) to the center of gravity is 4m. So, the force required is:
Force = 2400Nm / 4m = 600N
Hence, a force of 600N is required to begin lifting the end A of the pole.