Question

question 22
4 points save answer
(problem reference 4 - 1) a 23.0 kg apple crate accelerates along a frictionless, horizontal surface as the crate is pulled with an applied force of 14.5 n as shown in the drawing.
what is the normal force acting on the crate?
newtons apples
14.5 n
19°
a. 230 n
b. 225 n
c. 211 n
d. 220 n

Explanation:

Step1: Analyze vertical - forces

In the vertical direction, the forces acting on the crate are the gravitational force $F_g = mg$, the vertical component of the applied force $F_{y}=F\sin\theta$ (down - ward), and the normal force $N$. The net force in the vertical direction $F_{net,y}=0$ since there is no acceleration in the vertical direction. So, $N - mg+F\sin\theta = 0$.

Step2: Calculate gravitational force

The mass of the crate $m = 23.0\ kg$ and $g = 9.8\ m/s^{2}$. So, $F_g=mg=23.0\times9.8 = 225.4\ N$.

Step3: Calculate vertical component of applied force

The applied force $F = 14.5\ N$ and the angle $\theta = 19^{\circ}$. So, $F_{y}=F\sin\theta=14.5\times\sin(19^{\circ})\approx14.5\times0.326 = 4.727\ N$.

Step4: Solve for normal force

From $N - mg+F\sin\theta = 0$, we can express $N$ as $N=mg - F\sin\theta$. Substitute the values of $mg$ and $F\sin\theta$: $N = 225.4-4.727\approx220\ N$.

Answer:

d. 220 N