MOMENTS
A uniform rod 8m long weighing 5kg is supported horizontally by 2 vertical parallel strings @ p & Q and @ distances of 2m and 6m from one end.Weights of 1kg,1.5kg and 2kg are attached at distances of 1m, 5m and 7m respectively from the same end.
Find the tension in each vertical sting
Show all diagrams and workings
moments of Q Is equal to d distance of d weight times d weight plus d next moment dat is,5*2+1.5*3+2*5 = 1*6,
24.5Q and 6P
The tension in Q is 24.5 N AND P is 6N
To find the tension in each vertical string, we can use the principle of moments. The principle states that the sum of the clockwise moments must be equal to the sum of the anti-clockwise moments for the system to be in equilibrium.
First, let's draw a diagram of the rod and label the given information:
<----2m----> <---6m--->
|___________|_______|______|
O P Q
O represents the pivot point where the rod is supported.
P and Q are the positions where the strings are attached.
1kg, 1.5kg, and 2kg are weights attached at the distances of 1m, 5m, and 7m, respectively, from the same end.
Now, let's calculate the moments caused by the weights:
Moment due to 1kg weight:
- The weight of 1kg is acting downward at a distance of 1m from the pivot.
- So, the moment caused by this weight = 1kg * 1m = 1Nm (clockwise).
Moment due to 1.5kg weight:
- The weight of 1.5kg is acting downward at a distance of 5m from the pivot.
- So, the moment caused by this weight = 1.5kg * 5m = 7.5Nm (clockwise).
Moment due to 2kg weight:
- The weight of 2kg is acting downward at a distance of 7m from the pivot.
- So, the moment caused by this weight = 2kg * 7m = 14Nm (clockwise).
Next, let's calculate the moments caused by the rod itself:
The weight of the rod is 5kg, and its center of mass is at a distance of 4m from the pivot.
Moment due to the rod's weight:
- The weight of the rod is acting at its center of mass, which is 4m from the pivot.
- So, the moment caused by the rod's weight = 5kg * 4m = 20Nm (anti-clockwise).
To calculate the tensions in the strings, we need to find the net moment about the pivot point. Based on the principle of moments, the net moment should be zero.
Net moment = (Clockwise moments) - (Anti-clockwise moments)
Setting up the equation:
1Nm + 7.5Nm + 14Nm - 20Nm = 0
Simplifying the equation:
1Nm + 7.5Nm + 14Nm = 20Nm
22.5Nm = 20Nm
Now, let's find the tensions in the strings:
Let T1 be the tension in string P and T2 be the tension in string Q.
∑ Clockwise Moments = ∑ Anticlockwise Moments
(1kg * 1m * g) + (1.5kg * 5m * g) + (2kg * 7m * g) = (5kg * 4m * g) + (T1 * 2m * g) + (T2 * 6m * g)
Simplifying further:
9.81 * (1 + 7.5 + 14) = 20 * 9.81 + 2 * T1 * 9.81 + 6 * T2 * 9.81
Now, solve this equation to find the values of T1 and T2.
To find the tension in each vertical string, we can solve this problem using the principles of moments.
First, let's draw a diagram to visualize the given information:
```
|--------------------|
▼ ▼
P Q
▲ ▲
| 1m 3m 4m 7m 8m
```
Given:
Length of the rod (L) = 8m
Weight of the rod (W) = 5kg
Distances of strings from one end (p and q) = 2m and 6m
Weights attached at distances (1m, 5m, 7m) with masses (1kg, 1.5kg, 2kg)
Now, let's calculate the forces acting on the rod and balances the moments around point P and Q.
First, calculate the weight of the rod:
Weight of the rod (W) = mass × acceleration due to gravity
W = 5kg × 9.8m/s² = 49N
Next, let's calculate the moments around point P (using clockwise positive convention):
1. Moment due to the weight of the rod:
Moment around P due to the weight of the rod = Weight × distance
= 49N × 2m = 98Nm (clockwise)
2. Moment due to the attached weights:
Moment around P due to the weight at 1m = Mass × acceleration due to gravity × distance
Moment around P due to the weight at 1m = 1kg × 9.8m/s² × 1m = 9.8Nm (clockwise)
Moment around P due to the weight at 5m = 1.5kg × 9.8m/s² × 5m = 73.5Nm (clockwise)
Moment around P due to the weight at 7m = 2kg × 9.8m/s² × 7m = 137.2Nm (clockwise)
Now, let's calculate the moments around point Q (using clockwise positive convention):
1. Moment due to the weight of the rod:
Moment around Q due to the weight of the rod = Weight × distance
= 49N × 6m = 294Nm (clockwise)
2. Moment due to the attached weights:
Moment around Q due to the weight at 1m = 1kg × 9.8m/s² × 7m = 68.6Nm (counter-clockwise)
Moment around Q due to the weight at 5m = 1.5kg × 9.8m/s² × 3m = 44.1Nm (counter-clockwise)
Moment around Q due to the weight at 7m = 2kg × 9.8m/s² × 1m = 19.6Nm (counter-clockwise)
Now we can set up the equations based on the principle of moments:
Sum of clockwise moments = Sum of anticlockwise moments
For P:
98 + 9.8 + 73.5 + 137.2 = Tension in string P × distance p
319.5 = Tension in string P × 2m
For Q:
294 = Tension in string Q × distance q + 68.6 + 44.1 + 19.6
294 = Tension in string Q × 6m + 132.3
Solve these two equations using simultaneous equations to find the tension in string P and Q.
319.5 = Tension in string P × 2m
294 = Tension in string Q × 6m + 132.3
Now, solve these equations to find the tensions in each vertical string.