AR U. answered • 10/20/19

Experienced Physics and Math Tutor [Edit]

This question is solvable even without the figure (your referenced figure is missing :-)). I will split the problem into two parts:

1) bottom block, with mass m_{1} and 2) top block, with mass m_{2}

Draw the FBD for both cases and from that, we have the various components of the forces to be

bottom block

---------------------

x: ∑F_{1},_{x} = F_{1} - F_{s} = m_{1}a_{1,x --------------------------------------1}

y: ∑F_{1,y} = N_{1} - N_{2} - m_{1}g = 0 ==> N_{1} = N_{2} + m_{1}g --------------------------------------2

top block

--------------------

x: ∑F_{2},_{x} = F_{s} = m_{2}a_{2,x ------------------------------------- 3}

y: ∑F_{2,y} = N_{2} - m_{2}g = 0 ==> N_{2} = m_{2}g ------------------------------------- 4

b)

Assuming a_{2,x }= a_{1,x} = a, then from equation (3), you get

F_{s} = m_{2}a = µ_{s}N_{2} = µ_{s}m_{2}g ==> a = µ_{s}g = 0.43(9.8m/s^{2}) = 4.21m/s^{2}

a) From equation (1), we have

F_{1} = F_{s} + m_{1}a = µ_{s}g(m_{2} + m_{1}) = 0.43(9.8m/s^{2})(4.5kg + 15.0kg) = 82.17N

Your results (for the first case) are indeed correct :-)

============================================

Now let's do similar calculation for the case where the applied force is attached to the top block

Split the problem into two parts again: 1) bottom block and 2) top block

bottom block

-------------------------

x: ∑F_{1},_{x} = F_{s} = m_{1}a1_{,x ----------------------------------- 11}

_{y: } ∑F_{1,y} = N_{1} - N_{2} -m_{1}g_{ }= 0 ==> N_{1} = g(m_{2 }+ m_{1}) -----------------------------------22

top block

------------------------

x: ∑F_{2},_{x} = F_{2} - F_{s} = m_{2}a_{2},_{x ------------------------------------33}

y: ∑F_{2},_{y} = N_{2} - m_{2}g + 0 ==> N_{2} = m_{2}g ------------------------------------44

b)

Assuming a_{2},_{x} = a_{1,x} = a, then from equation (11), you get

F_{s} = m_{1}a = µ_{s}m_{2}g ==> a = µ_{s}g(m_{2}/m_{1}) = 0.43(9.8m/s^{2})(4.5kg/15.0kg) = 1.26m/s^{2}

a) From equation (33), you have

F_{2} = F_{s} + m_{2}a = µ_{s}m_{2}g + m_{2}a = µ_{s}m_{2}g + m_{2}µ_{s}g(m_{2}/m_{1}) = m_{2}µ_{s}g(1 + m_{2}/m_{1})

= 4.5kg(0.43)(9.8m/s^{2})(1+4.5kg/15.0kg) = 24.65N

-------------------------------------------------------------------------------------------------

F_{s}: static friction

a_{1}: acceleration of bottom (large) block

a_{2}: acceleration of top (small) block

µ_{s}: coefficient of static friction

g: acceleration due to gravity = 9.8m/s^{2}

N_{1}: normal force on bottom (large) block

N_{2}: normal force on top (small) block