If the water displaced by an object has a volume of 0.1 m3, what is the buoyant force exerted on that object?
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Answer:
X component of force is 60.62 N.
Y component of force is 35 N.
Force of gravity on the dog is 245 N.
Magnitude of normal force is 210 N.
Explanation:
Given:
Force of pull is,
Mass of the dog is,
Angle of inclination is, °
Acceleration due to gravity is,
The free body diagram of the dog is shown below.
The X and Y components of force of pull is given as:
Therefore, the X and Y components of the force are 60.62 N and 35 N respectively.
Force of gravity on the dog is the product of its mass and acceleration due to gravity. Thus,
Therefore, the force of gravity on the dog is 245 N.
Now, consider the motion in the vertical direction of the dog. As there is no motion in the vertical direction, the net force along the Y direction is 0. In other words, the total Upward force is equal to the total downward force.
From the free body diagram,
Therefore, the normal force acting on the dog is 210 N.
