A pole AB of length 10.0m and weight 600N has its center of gravity 4.0m from the end A, and lies on horizontal ground .Calculat
1 answer:
Answer:
The force required to begin to lift the pole from the end 'A' is 240 N
Explanation:
The given parameters for the pole AB are;
The length of the pole, l = 10.0 m
The weight of the pole, W = 600 N ↓
The distance of the center of gravity of the pole from the side 'A' = 4.0 m
Let '' represent the force required to begin to lift the pole from the end 'A' and let a force applied in the upwards direction be positive
For equilibrium, the sum of moment about the point 'B' = 0, therefore, taking moment about 'B', we have
× 10.0 m - W × 4.0 m = 0
∴ × 10.0 m = W × 4.0 m = 600 N × 4.0 m
× 10.0 m = 600 N × 4.0 m
∴ = 600 N × 4.0 m/(10.0 m) = 240 N
The force required to begin to lift the pole from the end 'A', = 240 N.
You might be interested in
Answer:
The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.
Explanation:
Given that,
Initial velocity u= 128 ft/sec
Equation of height
....(I)
(a). We need to calculate the maximum height
Firstly we need to calculate the time
From equation (I)
Now, for maximum height
Put the value of t in equation (I)
(b). The number of seconds it takes the object to hit the ground.
We know that, when the object reaches ground the height becomes zero
Hence, The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.