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A chair of weight 70.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 41.0 N directed at an angle of 37.0 ∘ below the horizontal and the chair slides along the floor. Part A Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair. Express your answer in newtons.

Answer :

opudodennis

Answer:

94.67 N

Explanation:

Consider a free body diagram with force, F of 41 N applied at an angle of 37 degrees while the weight acts downwards. Resolving the force into vertical and horizontal components, we obtain a free body diagram attached.

At equilibrium, normal reaction is equal to the sum of the weight and the vertical component of the force applied. Applying the condition of equilibrium along the vertical direction.  

[tex]\begin{array}{l}\\\Sigma {F_y} = 0\\\\N - W - F\sin \theta = 0\\\\N = W + F\sin \theta \\\end{array}[/tex]

Substituting 70 N for W, 41 N for F and [tex]\theta[/tex] for 37 degrees

N=70+41sin37=94.67441595  N and rounding off to 2 decimal places

N=94.67 N

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${teks-lihat-gambar} opudodennis

The magnitude of the normal force that the floor exerts on the chair is 94.6 N

The free-body diagram is attached below showing the resolved forces

The force F of 41 N is applied at an angle of 37 degrees

equilibrium condition:

At equilibrium, the normal reaction N is equal to the sum of the weight W and the vertical component of the force applied.

The wight W = mg = 70N

The normal force N is given by:

N = Fsin37 + mg

N = (41 × 0.6 + 70)N

N = 94.6 N is the normal force that the chair exerts.

Learn more about normal force:

https://brainly.com/question/2287912?referrer=searchResults

${teks-lihat-gambar} ConcepcionPetillo

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