A 5.00-kg block is placed on top of a 10.0-kg block (Fig. P5.44). A horizontal force of 45.0 N is applied to the 10-kg block, and the 5-kg block is tied to the wall. The coefficient of kinetic friction between all moving surfaces is 0.200. (a) Draw a free-body diagram for each block and identify the action-reaction forces between the blocks. (b) Determine the tension in the string and the magnitude of the acceleration of the 10-kg block

I got,
top block
a = (um1-t1)/m1

bottom block
a = (45 - 2u(m2)g)/m2

since they share acceleration i can just make them equal and solve for T1

Question

A 5.00-kg block is placed on top of a 10.0-kg block (Fig. P5.44). A horizontal force of 45.0 N is applied to the 10-kg block, and the 5-kg block is tied to the wall. The coefficient of kinetic friction between all moving surfaces is 0.200. (a) Draw a free-body diagram for each block and identify the action-reaction forces between the blocks. (b) Determine the tension in the string and the magnitude of the acceleration of the 10-kg block

I got,
top block
a = (um1-t1)/m1

bottom block
a = (45 - 2u(m2)g)/m2

since they share acceleration i can just make them equal and solve for T1

australopithecus · Answer

Here iss the free body diagram

also I can solve for a for the bottom block easily

australopithecus · Answer

can someone check my answer?

australopithecus · Answer

u = frictional coefficient

anonymous · Answer

in ur FBD, u forgot to consider the normal forces between the surfaces in contact.

anonymous · Answer

the FBD for the lower block will be:
|dw:1343133384894:dw|
where,
Rg = Normal reaction from the ground
Fr1 = friction between lower block and ground
N = Normal reaction from the upper block.