A syringe has an area of 1.4 cm2 at its barrel and then narrows down to an area of 0.08 mm2 at the needle end. If a force of 5.5 N is applied to the syringe, what is the force produced at the tip of the needle?
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To find the force produced at the tip of the needle, we can use the principle of conservation of energy, specifically the principle of continuity. According to this principle, the product of the area of the syringe at any point and the velocity of the fluid remains constant along the flow.
We can solve this problem using the equation:
A1 × v1 = A2 × v2
Where A1 and A2 are the areas at the barrel and needle end respectively, and v1 and v2 are the velocities of the fluid at those points.
First, let's convert the area of the needle end from mm^2 to cm^2 to make the units consistent:
0.08 mm^2 = 0.008 cm^2
Now, we can rearrange the equation to solve for the velocity at the needle end:
v2 = (A1 × v1) / A2
v2 = (1.4 cm^2 × v1) / 0.008 cm^2
Next, we need to find the velocity at the barrel end. Assuming the fluid is incompressible, the velocity at the barrel end can be calculated using the equation of motion:
F1 = A1 × P1
Where F1 is the applied force and P1 is the pressure at the barrel end.
P1 = F1 / A1
P1 = 5.5 N / 1.4 cm^2
P1 = 3.93 N/cm^2
Now, we can find the velocity at the barrel end using Bernoulli's equation:
P1 + (0.5 × ρ × v1^2) = P2 + (0.5 × ρ × v2^2)
since the syringe is at the same height, the potential energy term cancels out:
(0.5 × ρ × v1^2) = (0.5 × ρ × v2^2)
where ρ represents the density of the fluid.
Simplifying the equation and substituting the value of P1:
0.5 × ρ × v1^2 = 3.93 N/cm^2 + 0.5 × ρ × v2^2
At this point, we don't have enough information to calculate the velocities, so we can't solve the equation. To proceed further, we need either the density of the fluid or additional information.
Please provide more information if available, or provide the density of the fluid if known.
To find the force produced at the tip of the needle, we can use the principle of Pascal's Law, which states that when a pressure is applied to an enclosed fluid, the pressure is transmitted uniformly to all parts of the fluid.
Pascal's Law can be expressed as:
Pressure1/Area1 = Pressure2/Area2
In this case, we can consider the pressure at the barrel as Pressure1 and the pressure at the needle end as Pressure2.
Let's start by converting the given areas to the same unit, either cm^2 or mm^2. Since the force is given in newtons (N), we'll convert both areas to cm^2.
Area at the barrel = 1.4 cm^2
Area at the needle end = 0.08 mm^2
To convert mm^2 to cm^2, we need to divide by 100 (since 1 cm = 10 mm).
Area at the needle end = 0.08 mm^2 ÷ 100 = 0.0008 cm^2
Now, we can use Pascal's Law to calculate the pressure at the barrel and needle end.
Pressure at the barrel/Area at the barrel = Pressure at the needle end/Area at the needle end
Pressure at the needle end = (Pressure at the barrel/Area at the barrel) × Area at the needle end
To find the pressure at the barrel, we need to divide the force applied by the area at the barrel.
Pressure at the barrel = Force applied / Area at the barrel = 5.5 N / 1.4 cm^2
Now, we can substitute the values into the equation to find the pressure at the needle end.
Pressure at the needle end = (5.5 N / 1.4 cm^2) × 0.0008 cm^2
Finally, to find the force produced at the tip of the needle, we multiply the pressure at the needle end by the area at the needle end.
Force at the tip of the needle = Pressure at the needle end × Area at the needle end
By substituting the values, we can calculate the force produced at the tip of the needle.