Question

A 7.00-kg block is stacked on top of the 2.90-kg block. What is the magnitude Fof the force, acting horizontally on the 2.90-kg block

as before, that is required to make the two blocks start to move together?

Answer 1

F = 19.42 [N]

Step-by-step explanation:

In order to solve this problem, we must first make a free body diagram and determine the forces acting on each body. In the attached image we can see the respective free body diagrams with the respective equilibrium equations.

In order to determine the value of the force, a static analysis must be performed on the block of 2.9 [kg] so that the value of the Force F can be determined.

A coefficient of friction value between the block of 2.9 [kg] and the ground must be taken into account in order to find the force F. a coefficient of friction value of 0.2 shall be taken

By the sum of forces on the Y-axis equal to zero, the value of the normal force N can be determined on the block of 2.9 [kg].

Then knowing that the frictional force is equal to the product of the normal force by the coefficient of friction, we can determine the frictional force.

By the sum of forces on the X-axis equal to zero, we can determine the value of the Force F, as seen in the equations posed in the attached image.

Note: the value of the Force F depends on the value of the coefficient of friction between surfaces.