Question

A chair of weight 80.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 44.0 N directed at an angle of 43.0 ∘ 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.

Answer 1

The magnitude of the normal force that the floor exerts on the chair is about 110 Newtons

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Further explanation

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

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Given:

direction of force = θ = 43°

magnitude of force = F = 44.0 N

weight of chair = w = 80.0 N

Unknown:

magnitude of the normal force = N = ?

Solution:

We will solve this problem by using Newton's Law of Motion as follows:

$\Sigma F_y = 0$

$N - F \sin \theta - w = 0$

$N = F \sin \theta + w$

$N = 44.0 \sin 43.0^o + 80.0$

$N \approx 110 \texttt{ Newtons}$

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

Answer 2

The chair does not move along the vertical direction: this means that the net force on the vertical direction is zero.

There are three forces acting along the vertical direction:

- The weight, acting downward: $W=mg=80.0 N$

- The component of the force F perpendicular to the floor, acting downward: $F_y = F sin 43.0^{\circ}=(44.0 N)(sin 43.0^{\circ})=30.0 N$

- The normal force N, acting upwards.

Since the net force must be zero, we have:

$W+F_y-N=0$

From which we can find the value of N:

$N=W+F_y=80.0 N+30.0 N=110.0 N$