Ice Cube Pressure on Incline: Solve for F

[brncsfns5621]

An ice cube, p = 917 kg/m^3, is sliding down a 16 degree frictionless incline. The sides of the cube each have a length of .75 m. What pressure does the cube exert on the incline?

P = F / A
= F / .75^3
= F / .42


F = ma
= 917 * a

I am having a problem figuring out the Force for this equation. I know it's simple, just can't put my finger on it. Please help.

 

[radou]

Which force is acting on the cube? One component of that force is in the direction of the motion of the cube, and one is perpendicular to the contact surface (m^2 !) of the cube and the incline.

 

[kyle8921]

To solve for the force (F), we need to use Newton's second law of motion, which states that force (F) equals mass (m) multiplied by acceleration (a). In this case, the mass of the ice cube is given as 917 kg/m^3 and the acceleration is due to gravity, which is 9.8 m/s^2. Therefore, the force (F) can be calculated as:

F = m * a
= 917 kg/m^3 * 9.8 m/s^2
= 8986.6 N

Now, to find the pressure (P) exerted by the ice cube on the incline, we can use the formula P = F / A, where A is the area of contact between the ice cube and the incline. Since the cube is sliding down the incline, the area of contact is equal to the length of one side of the cube, which is 0.75 m. Therefore, the pressure exerted by the ice cube on the incline is:

P = 8986.6 N / 0.75 m^2
= 11,982.1 Pa

Therefore, the ice cube exerts a pressure of 11,982.1 Pa on the incline as it slides down. It is important to note that this calculation assumes a frictionless incline and does not take into account any external forces acting on the ice cube.