Question

a chair of mass 13.5 kg is sitting on the horizontal floor, the floor is not frictionless. you push on the chair with a force f = 45.0 n that is directed at an angle of 34.0° below the horizontal and the chair slides along the floor. use newton’s laws to calculate the normal force that the floor exerts on the chair. express your answer in newtons.

Explanation:

Step1: Find the weight of the chair

The weight \( W \) of an object is given by \( W = mg \), where \( m = 13.5\space kg \) and \( g = 9.8\space m/s^2 \).
\( W = 13.5\times9.8 = 132.3\space N \)

Step2: Find the vertical component of the applied force

The applied force \( F = 45.0\space N \) is at an angle \( \theta = 34.0^\circ \) below the horizontal. The vertical component \( F_y \) is \( F\sin\theta \) (since it's below the horizontal, it adds to the weight for the normal force).
\( F_y = 45.0\times\sin(34.0^\circ) \)
\( \sin(34.0^\circ) \approx 0.5592 \)
\( F_y = 45.0\times0.5592 \approx 25.164\space N \)

Step3: Calculate the normal force

In the vertical direction, the normal force \( n \) balances the weight and the vertical component of the applied force (since there's no acceleration in the vertical direction, \( \sum F_y = 0 \)). So \( n = W + F_y \).
\( n = 132.3 + 25.164 = 157.464\space N \approx 157\space N \) (or more precisely, we can calculate with more accurate steps)

Wait, let's recalculate with more precision:

First, \( W = 13.5\times9.81 = 132.435\space N \) (using \( g = 9.81\space m/s^2 \) for more accuracy)

\( F_y = 45.0\times\sin(34^\circ) \)
\( \sin(34^\circ) \approx 0.5591929035 \)
\( F_y = 45.0\times0.5591929035 \approx 25.16368066\space N \)

Then \( n = 132.435 + 25.16368066 = 157.5986807\space N \approx 158\space N \) (or depending on the value of \( g \) used, if \( g = 9.8 \), then \( W = 13.5\times9.8 = 132.3 \), \( F_y = 45\times\sin(34^\circ)\approx45\times0.5592 = 25.164 \), \( n = 132.3 + 25.164 = 157.464 \approx 157\space N \))

Let's check the standard approach:

In vertical equilibrium (since the chair doesn't move up or down), the normal force \( n \) is equal to the weight of the chair plus the vertical component of the applied force (because the applied force has a downward vertical component).

So \( n = mg + F\sin\theta \)

Substituting the values:

\( m = 13.5\space kg \), \( g = 9.8\space m/s^2 \), \( F = 45.0\space N \), \( \theta = 34.0^\circ \)

\( n = 13.5\times9.8 + 45.0\times\sin(34.0^\circ) \)

Calculate \( 13.5\times9.8 = 132.3 \)

\( \sin(34.0^\circ) \approx 0.5592 \), so \( 45.0\times0.5592 = 25.164 \)

Then \( n = 132.3 + 25.164 = 157.464 \approx 157\space N \) (or if we use \( g = 9.81 \), \( 13.5\times9.81 = 132.435 \), \( 45\times\sin(34^\circ)\approx25.164 \), \( n = 132.435 + 25.164 = 157.599 \approx 158\space N \))

Answer:

\( \boxed{157} \) (or \( \boxed{158} \) depending on \( g \) value; if \( g = 9.8 \), 157 N; if \( g = 9.81 \), 158 N)