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2 years ago

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A chair of mass 12.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force F

= 40.0 N that is directed at an angle of 37.0∘ below the horizontal, and the chair slides along the floor. (a) Draw a clearly labeled free-body diagram for the chair. (b) Use your diagram and Newton’s laws to calculate the normal force that the floor exerts on the chair.

1 answer:

Tanya [424]2 years ago

3 0

Answer: Normal force, N = 141.64 Newton

Explanation:

All the forces acting on the system and described in free body diagram are:

1) gravitational pull in downward direction

2) Normal force in upward direction

3) External force of 40 N acting at an angle of 37° with the horizontal can be resolved in two rectangular components:

i) F Cos 37° along the horizontal plane in forward direction and

ii) F Sin 37° along the vertical plane in downward direction

Applying the Newton's second law, net forces in the vertical plane are:

Net force, f = N - (mg + F Sin 37°)

As there is no acceleration in the vertical plane hence, net force f = 0.

So,

N - (mg + F Sin 37°) = 0

Adding (mg + F Sin 37°) both the sides in above equation, we get

N = mg + F Sin 37°

N = 12 $\times$ 9.8 + 40 $\times$ 0.601 because (Sin 37° = 0.601)

N = 117.6 + 24.04

N = 141.64 Newton

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This question is incomplete. The full question was:

<em>A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s. </em>

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$Vf^2=Vo^2+2*a*d$

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