Question

3 boxes, having masses m1=5kg, m2=10kg and m3=15 kg are stacked "sideways" (i.e., in x-) on a surface with coefficient of kinetic friction 0.7 mu_k; an external force of magnitude 15 N pushes on the right-most box, making an angle of 30 degrees relative to the horizontal; a second force of magnitude 20 N pushes in the opposite direction on the left-most box, as a result of which all three accelerate uniformly. Determine the numerical value all of the forces acting on each of the boxes (including all normal forces between the boxes), as well as the amount the boxes have moved after a time of 7 seconds. Keep in mind that the normal force acting specifically on the box subject to the external force is changed by the external force...

please make a diagram

Answer 1

weight of each box, 5g , 10g, 15 g, pushes the boxes down and normal force by the floor on the each box

N₅ = 5g ; N₁₀ = 10g ; N₁₅ = 15g

The right side force 15 N has 2 components 15 Cos(30) = 12.99 N horizontal and 15Sin(30) = 7.5 N vertical.

Net force acting on the boxes = 20- 12.99 = 7.01 N

The vertical component acts on the right most box and accordingly its normal reaction

N₁₅ = 15g + 7.5 N and

the frictional force = (15*9.8 + 7.5) *0.7 = 108.15 N

co-efficient of friction u = 0.7

Frictional force = uN

total frictional force = 5*9.8 *0.7 + 10*9.8*0.7 + 108.15 = 515.14 N

The net force acting on the 3-boxes is less than the total frictional force and hence the boxes cannot move. Static friction will be higher than the kinetic friction.

Please check the values correctly.

When the boxes move left side box pushes the right side box with a force of ma and the right side boxes pushes the left side box with the same force and hence net reaction between the boxes is 0