Two identical boxes with mass 10kg sit on the horizontal floor of a truck. There is friction between one of the boxes and the floor, and no friction between the other box and the floor. The two boxes are tied together with a massless string. The truck begins to slow down with acceleration -3.75 m/s2.
If neither block slides, how large a coefficient of static friction must there be between the floor and the one box on the left (the one with the force of friction)?
To find the coefficient of static friction between the floor and the box with friction, we can use the concept of Newton's second law and the equilibrium of forces acting on the system.
Here are the steps to solve the problem:
1. Draw a free-body diagram for the system. Consider both boxes and the forces acting on them.
F_frictionL F_frictionR
-------------------- --------------------
| |
<---| M1 |--->
| |
|<--- T --->|
| |
v v
Box with Box without
friction friction
The forces acting on the box with friction (M1) are its weight (W1 = m1 * g), the normal force (N1) from the floor, and the static friction force (F_frictionL). The forces acting on the box without friction (M2) are its weight (W2 = m2 * g), and the tension in the string (T).
2. Write down the equation for the sum of forces along the horizontal direction (x-axis) for the box with friction (M1):
F_frictionL - T = m1 * (-a)
The acceleration (a) is negative because the truck is slowing down. The mass (m1) and acceleration (-a) have opposite signs.
3. Write down the equation for the sum of forces along the horizontal direction (x-axis) for the box without friction (M2):
T = m2 * (-a)
4. Since the system is in equilibrium, the tension in the string (T) is the same for both boxes:
T = F_frictionL
5. Substitute equation (4) into equations (2) and (3):
F_frictionL - F_frictionL = m1 * (-a)
F_frictionL = m2 * (-a)
6. Substitute the given values into the equation:
10 * (-a) = 10 * (-3.75)
Note that both masses are 10 kg.
7. Simplify and solve for the acceleration (a):
a = 3.75 m/s²
8. Finally, calculate the coefficient of static friction using the equation:
F_frictionL = μ * N1
Since the normal force (N1) is equal to the weight (W1 = m1 * g):
F_frictionL = μ * m1 * g
Rearranging, we get:
μ = F_frictionL / (m1 * g)
Substitute the known values:
μ = 10 * 3.75 / (10 * 9.8)
μ = 0.3837
Therefore, the coefficient of static friction between the floor and the box with friction must be approximately 0.3837.