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 solve this problem, we need to understand the forces acting on the two boxes.
First, let's consider the forces acting on the box without friction:
1. The weight of the box: This force acts vertically downward and has a magnitude of m*g, where m is the mass of the box and g is the acceleration due to gravity.
Since the mass of each box is 10 kg, the weight of this box is (10 kg) * (9.8 m/s^2) = 98 N.
2. The tension in the string: Since the boxes are tied together, the tension in the string pulls the box towards the right.
The tension in the string will be the same for both boxes since they are tied together.
We need to find the tension in the string.
Now let's consider the forces acting on the box with friction:
1. The weight of the box: Similar to the box without friction, the weight of this box is 98 N.
2. The frictional force: This force acts between the box and the floor, opposing the motion or tendency to slide.
We need to find the coefficient of static friction (μs) between the box and the floor.
Since the truck is slowing down with an acceleration of -3.75 m/s^2, the net force acting on the two-box system is:
Net Force = (mass of the two boxes) * (acceleration of the truck)
= (20 kg) * (-3.75 m/s^2)
= -75 N
Now, let's analyze the forces acting on the box with friction and set up the equations of motion:
1. The vertical forces should cancel each other out since the boxes are not moving vertically.
Weight of the box = Normal force from the floor
98 N = Normal force
2. The horizontal forces should also cancel each other out since the boxes are not sliding.
Frictional force = Tension in the string
μs * Normal force = Tension in the string
Now let's substitute the values we know into the equations:
μs * 98 N = Tension in the string ---(1)
Net Force = Tension in the string - Frictional force
-75 N = Tension in the string - μs * 98 N
-75 N + μs * 98 N = Tension in the string ---(2)
Since the tension in the string is the same in both equations, we can equate equations (1) and (2):
μs * 98 N = -75 N + μs * 98 N
Simplifying:
75 N = μs * 98 N
Dividing both sides by 98 N:
μs = 75 N / 98 N
Calculating this value gives us:
μs ≈ 0.765
Therefore, the coefficient of static friction between the floor and the box with friction must be approximately 0.765.