How do you know that forces are balanced when static friction acts on an object?
1 answer:
By looking at the acceleration of the object.
In fact, Netwon's second law states that the resultant of the forces acting on an object is equal to the product between the mass m of the object and its acceleration:
So, when static friction is acting on the object, if the object is still not moving we know that all the forces are balanced: in fact, since the object is stationary, its acceleration is zero, and so the resultant of the forces (left term in the formula) must be zero as well (i.e. the forces are balanced).
In fact, Netwon's second law states that the resultant of the forces acting on an object is equal to the product between the mass m of the object and its acceleration:
So, when static friction is acting on the object, if the object is still not moving we know that all the forces are balanced: in fact, since the object is stationary, its acceleration is zero, and so the resultant of the forces (left term in the formula) must be zero as well (i.e. the forces are balanced).
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Answer:
a) Check Explanation
b) Check Explanation
c) The rate at which water is being pumped into the tank = 2.631 cm³/min
Explanation:
Let the rate of flow of water into the tank be k cm³/min
a) The image of the conical tank is presented in the attached image
Note, the radius and height of a cone are related through the similar triangles principle.
As shown in the attached image, it is evident that
r/h = 3/10
r = 3h/10 = 0.3 h
b) The quantities given in the problem.
- Shape of the tank, conical tank, Hence volume of the tank = πr²h/3
- total height of the tank, H = 10 cm
- Radius of the tank at the top, R = D/2 = 6/2 = 3 cm
- rate at which water is leaking from the tank = 1.5 cm³/min
- water is being pumped into the tank at constant rate of k cm³/min
- As at height of water, h = 2 cm, the rate of rise in water level = 1 cm/min
c) volume of the tank at any time = πr²h/3
Rate of change in the volume of water in the tank = (rate of flow into the tank) - (Rate of water flow out of the tank)
dV/dt = k - 1.5
V = πr²h/3 and r = 0.3 h, r² = 0.09 h²
V = 0.03πh³
dV/dt = (dV/dh) × (dh/dt)
dV/dh = 0.09π h²
dV/dt = 0.09π h² (dh/dt)
dV/dt = k - 1.5
0.09π h² (dh/dt) = k - 1.5
But at h = 2 cm, (dh/dt) = 1.0 cm/min
0.09π h² (dh/dt) = k - 1.5
0.09π 2² (1) = k - 1.5
k - 1.5 = 1.131
k = 1.5 + 1.131 = 2.631 cm³/min
