2D Force Problems

Question

A 50-N applied force (pulling 30 degrees up from the horizontal) accelerates a 5 kg box across a horizontal floor.

a. What is the normal force on the box?

b. If the coefficient of friction between the floor and box is 0.15, what is the acceleration of the box?

Answer 1

50N force acting 30 degrees up from horizontal would have 2 components: in the x and y directions

 

F_y = 50 * sin(30) = 25N

F_x= 50 * cos(30) = sqrt(3) / 2 = 43.3N

 

Weight acting down = 5 * 9.8 = 49N

Sum of forces in y direction must be zero if box is moving along horizontal,

 

25+normal force = 49

 

a) normal force = 49-25 = 24N

b)

 

Friction force = coefficient of friction * normal force = 24 * 0.15 = 3.6N

 

sum of forces in x direction =

43.3 - 3.6 = 39.7 = ma

    39.7 / 5 = 7.94 m / sec / sec