A UNIFORM PLANK OF MASS 40KG AND LENGTH 30m REST HORIZONTALLY ON TWO SUPPORT P AND Q, WHERE LENGTH PX IS 8m AND LENGTH QX IS 18 m. OBJECTS OF MASSES 10kg AND 5 kg ARE HANGED AT POINT M and N RESPECTIVELY, WHERE LENGTH MX IS 10 m and LENGTH NY IS 6m. IF THE SYSTEM REMAINS IN EQUILIBRIUM UNDER THE ACTION OF THREE FORCES, CALCULATE THE PRESSURE EXERTED ON I) P II) Q
Well, that's quite the balancing act! Let's see if we can calculate the pressure exerted on the supports.
To determine the pressure at each support point, we need to consider the torques acting about those points.
For support point P:
The torques acting on P are:
1. Weight of the uniform plank: Since the plank is in equilibrium, the weight of the plank must be balanced by the combined weights of the masses at M and N. The torque due to the plank's weight will be zero about P.
2. Weight of the object at M: The torque due to the weight at M is calculated by multiplying the weight (mass * gravity) by the perpendicular distance from P to M (8m). Let's call this torque MP.
3. Weight of the object at N: The torque due to the weight at N is calculated by multiplying the weight (mass * gravity) by the perpendicular distance from P to N (18m). Let's call this torque NP.
Since the system is in equilibrium, the sum of the torques acting on P must be zero:
Torque MP + Torque NP = 0
Similarly, for support point Q:
The torques acting on Q are:
1. Weight of the uniform plank: As mentioned earlier, the torque due to the plank's weight will be zero about Q.
2. Weight of the object at M: The torque due to the weight at M is calculated by multiplying the weight by the perpendicular distance from Q to M (28m). Let's call this torque MQ.
3. Weight of the object at N: The torque due to the weight at N is calculated by multiplying the weight by the perpendicular distance from Q to N (12m). Let's call this torque NQ.
Again, since the system is in equilibrium, the sum of the torques acting on Q must be zero:
Torque MQ + Torque NQ = 0
Now, let's calculate the torques and pressures:
1. Torque MP = (mass at M * gravity) * distance MP
= (10kg * 9.8m/s^2) * 8m
= 784 Nm
2. Torque NP = (mass at N * gravity) * distance NP
= (5kg * 9.8m/s^2) * 18m
= 882 Nm
3. Torque MQ = (mass at M * gravity) * distance MQ
= (10kg * 9.8m/s^2) * 28m
= 2744 Nm
4. Torque NQ = (mass at N * gravity) * distance NQ
= (5kg * 9.8m/s^2) * 12m
= 588 Nm
Now, let's calculate the pressures:
I) Pressure exerted on P:
Sum of torques about P = Torque MP + Torque NP = 784 Nm + 882 Nm = 1666 Nm
Pressure on P = (Sum of torques about P) / Distance PX
= 1666 Nm / 8m
= 208.25 N/m^2
II) Pressure exerted on Q:
Sum of torques about Q = Torque MQ + Torque NQ = 2744 Nm + 588 Nm = 3332 Nm
Pressure on Q = (Sum of torques about Q) / Distance QX
= 3332 Nm / 18m
= 185.11 N/m^2
So, the pressure exerted on support P is 208.25 N/m^2, and the pressure exerted on support Q is 185.11 N/m^2.
Now, keep in mind that these calculations assume the uniform plank, and the hanging objects are massless. Otherwise, we would have to take into account their weights in the calculations as well.
To calculate the pressure exerted on support points P and Q, we need to consider the equilibrium of the system.
For equilibrium, the sum of the clockwise moments must equal the sum of the anticlockwise moments.
Let's calculate the moments for each force separately.
1) Force at point M (10 kg):
The weight of the 10 kg object exerts a force of 10 * 9.8 = 98 N.
The distance between the force and point P is 10 m. So the moment about P is 98 * 10 = 980 Nm (clockwise).
2) Force at point N (5 kg):
The weight of the 5 kg object exerts a force of 5 * 9.8 = 49 N.
The distance between the force and point Q is 6 m. So the moment about Q is 49 * 6 = 294 Nm (anticlockwise).
To maintain equilibrium, the resultant moment about any point must be zero.
Now let's calculate the resultant moments about point P:
The moment at P due to the weight of the uniform plank is zero because it is acting at the center of mass.
So the total moment about P would be only the moment due to the force at point M: 980 Nm (clockwise).
For equilibrium, the sum of clockwise and anticlockwise moments must be equal, so the total moment about Q would be equal to the moment at P. Therefore, the total moment about Q would also be 980 Nm (clockwise).
Now, let's calculate the pressure exerted on P and Q:
1) Pressure exerted on point P:
The total weight on the plank is the sum of the mass of the plank (40 kg), the 10 kg object, and the 5 kg object:
Total weight = (40 + 10 + 5) * 9.8 = 55 * 9.8 = 539 N.
The pressure exerted on point P would be the total weight divided by the contact area at point P, which is just the length of PX times the width of the plank:
Pressure on P = 539 N / (8 m * width of the plank).
2) Pressure exerted on point Q:
As explained earlier, the total moment about Q is 980 Nm (clockwise).
To maintain equilibrium, the sum of clockwise and anticlockwise moments about Q must be equal. So, the moment about Q due to the weight of the uniform plank is 980 Nm (anticlockwise).
The pressure exerted on point Q would be the total weight (539 N) minus the weight of the uniform plank (40 kg * 9.8) divided by the contact area at point Q, which is just the length of QX times the width of the plank:
Pressure on Q = (539 N - 40 kg * 9.8) / (18 m * width of the plank).
Note: The width of the plank is not provided in the given information. You would need to know the width to calculate the exact pressures on points P and Q.
To calculate the pressure exerted on support points P and Q, we first need to determine the forces acting at those points.
Let's analyze the forces acting on the plank:
1) Gravity: The weight of the plank itself (40 kg) acts vertically downward at its center of mass. The weight can be calculated using the formula: weight = mass × gravity, where gravity is approximately 9.8 m/s^2.
Weight of the plank (P) = 40 kg × 9.8 m/s^2 = 392 N
2) Forces from the objects: The objects at points M and N exert downward forces on the plank due to their weights. We can calculate these forces using the same formula:
Force at M = 10 kg × 9.8 m/s^2 = 98 N
Force at N = 5 kg × 9.8 m/s^2 = 49 N
Now, let's consider the equilibrium condition:
For a uniform plank in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.
Clockwise moments:
Moment at P = Force at M × Distance PM = 98 N × 10 m = 980 Nm
Moment at Q = Force at N × Distance QN = 49 N × 6 m = 294 Nm
Anticlockwise moment:
Moment due to the weight of the plank at P = Weight of the plank × Distance PX = 392 N × 8 m = 3136 Nm
Moment due to the weight of the plank at Q = Weight of the plank × Distance QX = 392 N × 18 m = 7056 Nm
To maintain equilibrium, the sum of the clockwise and anticlockwise moments must be equal:
980 Nm + 294 Nm = 3136 Nm + 7056 Nm
1274 Nm = 10192 Nm
Since the above equation is not balanced, there is an error in the question or its data. Please double-check the given information and try again.