Question

2.m limiting cases
quantitative analysis
part c: derive expressions for the following in terms of m, μ, f, and θ:
i. the force that the ground exerts on the system.
ii. the force of friction the ground exerts on the system.
iii. the acceleration of the system.

Explanation:

Step1: Analyze vertical - force equilibrium for normal force

In the vertical direction, the sum of forces is zero. Let the normal force exerted by the ground on the system be $N$. The weight of the object is $mg$ and the vertical component of the applied force is $F_y\sin\theta$. So, $\sum F_y = N - mg - F_y\sin\theta=0$.
$N = mg + F_y\sin\theta$

Step2: Calculate frictional - force

The force of friction $f$ is given by the formula $f=\mu N$. Substituting $N = mg + F_y\sin\theta$ into the formula for friction, we get $f=\mu(mg + F_y\sin\theta)$

Step3: Apply Newton's second law for acceleration

In the horizontal direction, assume the horizontal component of the applied force is $F_y\cos\theta$. According to Newton's second - law $\sum F_x=ma$. So, $F_y\cos\theta - f=ma$. Substituting $f=\mu(mg + F_y\sin\theta)$ into the equation, we have $F_y\cos\theta-\mu(mg + F_y\sin\theta)=ma$. Then, $a=\frac{F_y\cos\theta-\mu(mg + F_y\sin\theta)}{m}$

Answer:

i. $N = mg + F_y\sin\theta$
ii. $f=\mu(mg + F_y\sin\theta)$
iii. $a=\frac{F_y\cos\theta-\mu(mg + F_y\sin\theta)}{m}$