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Question: What normal force is exerted on a stationary flat plate held perpendicular to a jet of water? The ho...

What normal force is exerted on a stationary flat plate held perpendicular to a jet of water? The horizontal speed of the water is 80cm/s80cm/s and 30cm330c{{m}^{3}} of the water hits the plate each second. Assume that the water moves parallel to the plate after striking it. One cubic centimeter of water has a mass of one gram.
A. 0.03N0.03N
B. 0.024N0.024N
C. 0.04N0.04N
D. 0.036N0.036N

Explanation

Solution

We have learned that the net force acting on an object is equal to the change in momentum over time. This law can be applied here. With the given speed and the volume of water we can find out momentum of water. Substitute the value of momentum in the force - momentum equation to find out the force acting on the plate.
Formula used:
Momentum = mass × velocity\text{Momentum = mass }\times \text{ velocity}
Force = Final momentum - initial momentumtime interval\text{Force = }\dfrac{\text{Final momentum - initial momentum}}{\text{time interval}}

Complete answer:
Given,
Speed of water = 80cm/s = 0.8m/s\text{Speed of water = 80cm/s = 0}\text{.8m/s}
Volume of water hitting the plate = 30cm3\text{Volume of water hitting the plate = 30c}{{\text{m}}^{\text{3}}}
Mass of 1cm3water = 1g\text{Mass of 1c}{{\text{m}}^{\text{3}}}\text{water = 1g}
Then,
Mass of 30cm3water = 1×30 = 30g = 0.03Kg\text{Mass of 30c}{{\text{m}}^{\text{3}}}\text{water = 1}\times \text{30 = 30g = 0}\text{.03Kg}
We have,
Momentum = mass × velocity\text{Momentum = mass }\times \text{ velocity}
Here,
Velocity of water is 0.8m/s\text{Velocity of water is 0}\text{.8m/s}
Then,
Momentum of 0.03Kg water hitting the plate = 0.03×0.8 = 0.024 Kgm/s\text{Momentum of 0}\text{.03Kg water hitting the plate = 0}\text{.03}\times \text{0}\text{.8 = 0}\text{.024 Kgm/s}
Hence, Initial momentum = 0.024 Kgm/s\text{Initial momentum = 0}\text{.024 Kgm/s}
We have,
Force = Final momentum - initial momentumtime interval\text{Force = }\dfrac{\text{Final momentum - initial momentum}}{\text{time interval}}
Velocity of water after hitting the plate is zero as the water moves parallel to the plate after striking it.
Then,
Final momentum of water = 0 Kgm/s\text{Final momentum of water = 0 Kgm/s} And, time =1s\text{time =1s}
Therefore,
Force on plate = 0.024-01=0.024N\text{Force on plate = }\dfrac{\text{0}\text{.024-0}}{1}=0.024N
Normal force on plate, N= 0.024N\text{Normal force on plate, N= 0}\text{.024N}

So, the correct answer is “Option B”.

Additional Information:
A normal force is a contact force which surfaces exert to prevent objects from passing through each other. It is not applicable for two surfaces which are not in contact. It acts perpendicular to the surface. If normal force is the only force counteracting on a body, it is simply equal to the weight of the object. It is represented as NN. The normal force can be less than the weight of the object, if the object is on an inclined surface.

Note:
A normal force acts on a body even if it is at rest. The magnitude of the normal force can only be determined by analyzing the situation. The normal force acting on a body changes in response to other forces acting on the object. And normal force can never be zero.