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brilliants [131]
4 years ago
11

A chair of weight 95.0N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 3

9.0N directed at an angle of 37.0 degrees below the horizontal and the chair slides along the floor.
1)Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair.

Physics
2 answers:
AlladinOne [14]4 years ago
7 0

Answer:

Answer:

118.4 N

Explanation:

weight of chair, mg = 95 N

Push, F = 39 N

Ф = 37 ° below x axis

Let n be the normal force.

So, by using the diagram and resolve the components of Force F.

n = mg + F SinФ

n = 95 + 39 Sin 37°

n = 95 + 39 x 0.6

n = 118.4 N

Explanation:

mariarad [96]4 years ago
7 0

Answer:118.47

Explanation:

Given

force (F)= 39 N acting at an angle of 37 below horizontal

Force can be divided in to sin and cos component

Thus Fsin37 will act in Y direction

Total Force in Y direction is Normal reaction , Fsin37 and weight

N-mg-Fsin37=0

N=95+39\times \frac{3}{5}

N=118.47 N

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A disk 7.90 cm in radius rotates at a constant rate of 1 190 rev/min about its central axis. (a) Determine its angular speed. 12
Tanya [424]

Answer:

124.62\ \text{rad/s}

3.71\ \text{m/s}

1.23\ \text{km/s}^2

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Explanation:

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Angular speed is given by

\omega=N\dfrac{2\pi}{60}\\\Rightarrow \omega=1190\times \dfrac{2\pi}{60}\\\Rightarrow \omega=124.62\ \text{rad/s}

The angular speed is 124.62\ \text{rad/s}

r = 2.98 cm

Tangential speed is given by

v=r\omega\\\Rightarrow v=2.98\times 10^{-2}\times 124.62\\\Rightarrow v=3.71\ \text{m/s}

Tangential speed at the required point is 3.71\ \text{m/s}

Radial acceleration is given by

a=\omega^2r\\\Rightarrow a=124.62^2\times 7.9\times 10^{-2}\\\Rightarrow a=1226.88\approx 1.23\ \text{km/s}^2

The radial acceleration is 1.23\ \text{km/s}^2.

t = Time = 2.06 s

Distance traveled is given by

d=vt\\\Rightarrow d=\omega rt\\\Rightarrow d=124.62\times 7.9\times 10^{-2}\times 2.06\\\Rightarrow d=20.28\ \text{m}

The total distance a point on the rim moves in the required time is 20.28\ \text{m}.

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