Question

This is a problem about a child pushing a stack of two blocks along a horizontal floor. The masses of the blocks, and the coefficients of friction between the two blocks and between the lower block and the floor will be given. In order to do the pushing, the child will only be touching one of the two blocks. The mass of the upper block in the stack is 0.760 kg . The mass of the lower block in the stack is 1.630 kg . The coefficients of friction between the two blocks are: static 0.790, and kinetic 0.660. The child's mother, who likes to encourage his experiments, has oiled a small strip of the horizontal floor so that it is very slick; the coefficient of kinetic friction between the oiled section of floor and the lower block is only 0.080 and the coefficient of static friction is insignificantly different. Before the pushing starts, here is a question about the vertical forces acting on the two blocks.
Required:
What is the vertical component of the contact force on the lower block by the floor?

Answer 1

Answer:

 N = 23.4 N

Explanation:

After reading that long sentence, let's solve the question

The contact force is the so-called normal in this case we can find it by writing the translational equilibrium equation for the y axis

            N - w₁ -w₂ =

            N = m₁ g + m₂ g

            N = g (m₁ + m₂)

let's calculate

            N = 9.8 (0.760 + 1.630)

            N = 23.4 N

This is the force of the support of the two blocks on the surface.