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3 years ago

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A chair of weight 100 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 4

3.0 N directed at an angle of 35.0 degrees below the horizontal and the chair slides along the floor. Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair. Express your answer in newtons.

1 answer:

otez555 [7]3 years ago

7 0

Answer:

The normal force will be "122.8 N".

Explanation:

The given values are:

Weight,

W = 100 N

Force,

F = 40 N

Angle,

θ = 35.0°

As we know,

⇒ $N=W+FSin \theta$

On substituting the given values, we get

⇒ $= 100N+40N \ Sin \theta$

⇒ $=100N+22.8$

⇒ $=122.8 \ N$

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For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

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