Question

Two masses, m1 and m2, are falling but not freely. In addition to gravity, there is also a force F1 applied directly to m1 in the downward direction and a force F2 applied directly to m2 in the horizontal direction. Friction (µs) is present between the two masses and the forces are applied such that they do not rotate. The force F2 is as large as it can be and not have m2 slide relative to m1. (a) Find an expression for the acceleration of the center of mass of the m1 + m2 system in terms of m1, m2, F1, F2, and g? (b) Draw a FBD for each mass separately. Identify motion constraints and Newton's 3rd law force pairs. (c) Write down Newton's 2nd law applied to each mass separately. (d) If both masses are each 2 kg, the coefficient of static friction between the surface is µs = 1/2, and F1 = 25 N, What is the value of F2?

Answer 1

Two masses, m1 and m2, are falling but not freely.

(a) An expression for the acceleration of the center of mass of the m₁ + m₂ system in terms of m₁, m₂, F₁, F₂, and g which is given as :;

on the horizontal axis, force components is given as -

F_x = F₂                                              { eq.1 }

using newton's second law of motion,

(m₁ + m₂) a_12x = F₂

a_12x = F₂ / (m₁ + m₂)                                     { eq.2 }

on the vertical axis, the force components is given as :

F_y = F₁ + (m₁ + m₂) g                            { eq.3 }

using newton's second law of motion,

(m₁ + m₂) a_12y = F₁ + (m₁ + m₂) g   

a_12y = [F₁ + (m₁ + m₂) g] / (m₁ + m₂)   

a_12y = [F₁ + (m₁ + m₂)] + g                                              { eq.4 }

acceleration of the center of mass, a (x , y) is given as ::

a = ( F₂ / (m₁ + m₂) , [F₁ + (m₁ + m₂)] + g )       

(b) Free body diagram for each mass which can be shown below as ::

For mass m₁ -

For mass m₂ -

motion constraints of the system is given as :

= = =

Newton's 3rd law force pairs is given as :

= - =

and = -

(c) Applying to newton's second law of motion,

For mass m₁ on x-direction, F_x = F_N                             { eq.5 }

m₁ a_x = F_N     

For mass m₁ on y-direction, F_y = m₁ g + F₁ - F_N    { eq.6 }

inserting the values in eq.6,

m₁ a_y = m₁ g + F₁ - F_N

For mass m₂ on x-direction, F_x = F₂ - F_N                                 { eq.7 }

m₂ a_x = F₂ - F_N   

For mass m₂ on y-direction, F_x = m₂ g + F_N                                  { eq.8 }

m₂ a_y = m₂ g + F_N      

(d) Given that :

mass of each block, m = 2 kg

coefficient of static friction between the surface, _s = 0.5

Force, F₁ = 25 N   and

the value of F₂ is given as ::

using eq. m₁ a_y = m₁ g + F₁ - F_N { eq.9 }

where, F_N = m₁ a_x /

inserting the value of 'F_N' & 'a_y' in eq.9,

m₁ ([F₁ / (m₁ + m₂)] + g) = m₁ g + F₁ - (m₁ a_x / )

where, a_x = F₂ / (m₁ + m₂)

m₁ ([F₁ / (m₁ + m₂)] + g) = m₁ g + F₁ - m₁ [F₂ / (m₁ + m₂)] /

rearranging an above eq.

F₁ ([m₁ / (m₁ + m₂)] + g) = m₁ g + F₁ - [m₁ / (m₁ + m₂)] (F₂ / )    { eq.10 }

inserting the given values eq.10,

(25 N) ([2 kg / (2 kg + 2 kg)] + 9.8 m/s²) = (25 kg) (9.8 m/s²) + (25 N) - [2 kg / (2kg + 2kg)] (F₂ / 0.5)]

(25 N) (10.3 N) = (245 N) + (25 N) - (0.5) (F₂ / 0.5)

(257.5 N) = (270 N) - F₂

F₂ = 12.5 N