A uniform rod AB, 100cm long and of mass 2kg is supported horizontally at points P and Q, 40cm and 65cm respectively from A. A mass of 1kg is hung at a point 25cm from A. If the rod is in equilibrium, calculate the reactions at the support(Take g= 10m/s²).?
use P and Q individually as the center of rotation
sum the moments to find the reaction forces
To solve this problem, we can use the principles of torque and equilibrium.
First, let's calculate the weight of the rod and the weight of the 1kg mass hanging from the rod.
Weight of the rod (W1) = mass of the rod (m1) * gravitational acceleration (g)
= 2 kg * 10 m/s²
= 20 N
Weight of the 1kg mass (W2) = mass of the 1kg mass (m2) * gravitational acceleration (g)
= 1 kg * 10 m/s²
= 10 N
Next, let's calculate the torques acting on the rod.
Torque of W1 around point P (T1) = distance of point P from point A (D1) * W1
= 40 cm * 20 N
= 800 Ncm
Torque of W2 around point P (T2) = distance of the point 25cm from point A to point P (D2) * W2
= 25 cm * 10 N
= 250 Ncm
Torque of the reaction force at point Q around point P (T3) = distance of point Q from point P (D3) * reaction force at point Q (R)
= 25 cm * R
= 25R Ncm
Since the rod is in equilibrium, the sum of the torques acting on it should be zero.
Sum of torques = T1 - T2 - T3 = 0
Substituting the values we obtained:
800 Ncm - 250 Ncm - 25R Ncm = 0
Combining like terms:
550 Ncm - 25R Ncm = 0
Simplifying:
550 Ncm = 25R Ncm
R = 550 Ncm / 25 Ncm
Finally, converting Ncm to N:
R = 22 N
Therefore, the reaction force at point Q (R) is 22 N. At point P, the reaction force can be calculated by subtracting the weight of the rod and the 1kg mass from the sum of the reaction forces:
Reaction force at point P = (W1 + W2) - R
= 20 N + 10 N - 22 N
= 8 N
Therefore, the reaction force at point P is 8 N.
To calculate the reactions at the supports, we need to consider the forces acting on the rod in equilibrium.
First, let's denote the reaction at point P as RP and the reaction at point Q as RQ. Since the rod is in equilibrium, the sum of the forces acting on the rod in the vertical direction should be zero, and the sum of the moments (torques) acting on the rod about any point should also be zero.
Let's analyze the forces acting on the rod:
1. Weight of the rod (Mg): The weight of the rod acts vertically downwards and can be considered to act at its center of mass, which is at 50 cm from point A. So, the magnitude of the weight is 2 kg * 10 m/s² = 20 N.
2. Weight of the 1 kg mass (mg): The weight of the 1 kg mass acts vertically downwards and can be considered to act at a distance of 25 cm from point A. So, the magnitude of the weight is 1 kg * 10 m/s² = 10 N.
3. Reaction at point P (RP): The reaction at point P acts vertically upwards and can be considered to act at a distance of 40 cm from point A.
4. Reaction at point Q (RQ): The reaction at point Q acts vertically upwards and can be considered to act at a distance of 65 cm from point A.
Now, let's analyze the torques acting on the rod:
1. Torque due to the weight of the rod about point P: This torque is given by the product of the weight of the rod and the perpendicular distance from point P, which is 40 cm. So, the torque about point P is (20 N * 0.4 m) = 8 Nm.
2. Torque due to the weight of the 1 kg mass about point P: This torque is given by the product of the weight of the 1 kg mass and the perpendicular distance from point P, which is 25 cm. So, the torque about point P is (10 N * 0.25 m) = 2.5 Nm.
3. Torque due to the reaction at point Q about point P: This torque is given by the product of the reaction at point Q and the perpendicular distance from point P, which is (65 cm - 40 cm). So, the torque about point P is (RQ * 0.25 m) = 0.25RQ Nm.
Since the rod is in equilibrium, the sum of the torques about point P should be zero:
8 Nm + 2.5 Nm + 0.25RQ Nm = 0
Simplifying this equation, we get:
10.75 Nm + 0.25RQ Nm = 0
0.25RQ Nm = -10.75 Nm
RQ = -10.75 Nm / 0.25 m
RQ = -43 N
The negative sign indicates that RQ is acting downwards. However, in the given problem, the rod is supported horizontally, so the reaction at point Q cannot be negative. Therefore, we disregard this negative value.
Next, let's calculate the reaction at point P:
To find RP, we need to consider the vertical equilibrium of forces:
Sum of upward forces = Sum of downward forces
RP + RQ + Mg + mg = 0
RP + 0 + 20 N + 10 N = 0
RP = -30 N
Again, the negative sign indicates that RP is acting downwards, which is not possible in this case. Therefore, we disregard this negative value.
Since both RQ and RP cannot be negative, it suggests that the rod is not in equilibrium as stated in the problem. Please double-check the problem statement or provide additional information if required.