There is a 5kg box sitting on top of a 10kg box. The static friction between them is .5. What is the maximum tension that can be placed on the bottom box without causing the 5kg box to slip. The surface is flat.

max friction= .5*5g

max acceleration= maxfriction/mass
max tension= 15kg*accmax= 15(.5*5g)/5

To find the maximum tension that can be placed on the bottom box without causing the 5kg box to slip, we need to consider the forces acting on the system.

First, let's define the variables:
m1 = mass of the 5kg box = 5 kg
m2 = mass of the 10kg box = 10 kg
μ = coefficient of static friction between the boxes = 0.5

The maximum tension force (T) that can be applied to the bottom box without causing the 5kg box to slip can be determined by analyzing the frictional force between the boxes.

The equation for the maximum static frictional force (F_friction_max) between two objects can be calculated using the formula:

F_friction_max = μN

Where N is the normal force between the two objects. In this case, the normal force (N) is equal to the weight of the 5kg box (m1 * g), where g is the acceleration due to gravity (approximately 9.8 m/s^2).

N = m1 * g = 5 kg * 9.8 m/s^2 = 49 N

Substituting the value of N into the equation for the maximum static frictional force, we get:

F_friction_max = μ * N = 0.5 * 49 N = 24.5 N

Therefore, the maximum tension (T) that can be placed on the bottom box without causing the 5kg box to slip is equal to the maximum static frictional force (F_friction_max) between the boxes, which is 24.5 N.