5 books weigh 25 N
25*.2 = 5
a)1 N
b)5 N
c)3 N
d)7 N
25*.2 = 5
F(static friction) = coefficient of static friction * normal force
In this case, the normal force is the weight of the bottom book, since it is supporting the weight of the entire stack. The weight of one book is 5 N, so the weight of the bottom book is also 5 N.
F(static friction) = 0.2 * 5 N
F(static friction) = 1 N
Therefore, the horizontal force required to start sliding the top five books off the bottom one is 1 N.
So, the correct answer is:
a) 1 N
1. First, find the maximum force of static friction (Ffric) using the formula:
Ffric = µ * N
where µ is the coefficient of static friction and N is the normal force.
2. Since the books are stacked vertically, the normal force acting on the top five books is the weight of all six books combined. Calculate the normal force (N) using:
N = total weight of the books
3. Finally, compare the maximum force of static friction to the given options to determine the correct answer.
Let's go through the solution step by step:
1. Calculate the maximum force of static friction:
Ffric = µ * N
Ffric = 0.2 * N ... (equation 1)
2. Calculate the normal force (N):
Since each book weighs 5 N and there are six books, the total weight of the books is:
Total weight = 5 N/book * 6 books = 30 N
Therefore, the normal force acting on the top five books is also 30 N.
3. Plug the values into equation 1 to calculate the maximum force of static friction:
Ffric = 0.2 * 30 N
Ffric = 6 N
Now, compare the maximum force of static friction (6 N) with the given options:
a) 1 N
b) 5 N
c) 3 N
d) 7 N
Since the maximum force of static friction is 6 N, the answer is not a, b, or c. Thus, the correct answer is d) 7 N.