Answer:
Finding the Maximum Acceleration
Understanding the Problem
We're asked to find the maximum acceleration a car can have without causing a coffee cup on its roof to slide off. We're given the coefficient of static friction between the cup and the roof.
Solution
The maximum force of static friction (Fs) is given by:
* Fs = μs * N
Where:
* Fs is the force of static friction
* μs is the coefficient of static friction (0.23)
* N is the normal force (equal to the weight of the cup in this case)
Newton's second law states:
* F = ma
Where:
* F is the net force
* m is the mass of the object (the cup)
* a is the acceleration
Since the frictional force is the only force acting on the cup in the horizontal direction, we can equate the two equations:
* μs * N = ma
The normal force N is equal to the weight of the cup, which is mg, where g is the acceleration due to gravity (approximately 9.81 m/s²). Substituting this into the equation:
* μs * mg = ma
The mass of the cup (m) cancels out:
* μs * g = a
Now we can calculate the maximum acceleration:
* a = μs * g = 0.23 * 9.81 m/s² ≈ 2.25 m/s²
Therefore, the maximum acceleration the car can have without causing the cup to slide is 2.3 m/s² (to two significant figures).
Answer:
maximum acceleration = 2.3 m/s
Explanation:
To find the maximum acceleration the car can have without causing the cup to slide, we first create the free body diagram (see the attached picture).
As the cup does not move in both x and y directions, that means both [tex]\bf\Sigma F_x[/tex] and [tex]\bf\Sigma F_y[/tex] equal to 0. Now, we can calculate the acceleration using the force resultants.
[tex]\begin{aligned}\\\Sigma F_y&=0\\N-w&=0\\N&=w\\N&=mg\ \Longrightarrow\texttt{ since }w = mg\\\end{aligned}[/tex]
[tex]\texttt{Given the coeffiecient of static friction }(\mu_s)=0.23[/tex]
[tex]\begin{aligned}\\\Sigma F_x&=0\\F-f_{s(max)}&=0\\F&=f_{s(max)}\\ma_{(max)}&=\mu_s N \Longrightarrow\texttt{ since }F=ma\texttt{ and }f_{s(max)}=\mu_s N\\ma_{(max)}&=\mu_s mg\\a_{(max)}&=\mu_s g\\&=0.23\times9.8\\&=2.254\\&\approx\bf2.3\ m/s\end{aligned}[/tex]
