Physics
posted by April .
A block of mass M1=2.9 kg rests on top of a second block of mass2=5.5 kg, and the second block sits on a surface that is so slippery that the friction can be assumed to be zero.
Question:
(a) if the coefficient of static friction between the blocks is 0.21, how much force can be applied to the top block without the blocks slipping apart?
(b) how much force can be applied to the bottom block for the same result?

(a) 0.21*M1*g = 5.968 N
(b) The maximum friction force that M2 can apply to M1 without M1 slipping is also 5.968 N. That will cause M1 AND M2 to accelerate at rate
a = 5.968 N/M1 = 2.058 m/s^2
The force that would have to be applied to M2 to make this happen is
(M1 + M2)*a = 8.4 * 2.058 = 17.29 N 
The answer to part a above is incorrect. You must remember that the force of friction resisting the applied force on M1 also applies an equal and opposite force on M2. The applied force must overcome /both/ of these.
The sum of forces on the X axis for M2 is just this equal and opposite force (Force of friction). Setting this to ma will allow you to solve for an acceleration value.
You can use this acceleration value in solving for the applied force on M1. (T  Ffriction = ma).
Hope this helps. 
(a) Top block:
Ffriction=M1a
F=2.9a+(0.21*2.9*9.8)
Bottom block:
friction=M2a
a=(0.21*2.9*9.8)/5.5
F=(0.21*2.9*9.8*2.9)/5.5+(0.21*2.9*9.8)=9.12N
(b) Bottom block:
Ffriction=M2a
F=5.5a+(0.21*2.9*9.8)
Top block: 0.21*2.9*9.8=2.9a
a=0.21*9.8
F=(0.21*5.5*9.8)+(0.21*2.9*9.8)=17.29N
Respond to this Question
Similar Questions

Physics
A block with mass M rests on a frictionless surface and is connected to a horizontal spring of force constant k, the other end of which is attached to a wall. A second block with mass m rests on top of the first block. The coefficient … 
Physics (dynamics)
A 47.0 kg slab rests on a frictionless floor. A 18.0 kg block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.50, while the kinetic coefficient of friction is 0.40. The top block is … 
Physics
A block of mass m = 2.00 kg rests on the left edge of a block of mass M = 8.00 kg. The coefficient of kinetic friction between the two blocks is 0.300, and the surface on which the 8.00 kg block rests is frictionless. A constant horizontal … 
Physics
A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first block. The surfaces of the blocks are rough. A constant force of magnitude F is applied to the first block … 
physics
A block with mass = 5.0 rests on a frictionless table and is attached by a horizontal spring ( = 130N ) to a wall. A second block, of mass = 1.35 , rests on top of the block . The coefficient of static friction between the two blocks … 
physics
A block with mass = 5.0 rests on a frictionless table and is attached by a horizontal spring ( = 130N ) to a wall. A second block, of mass = 1.35 , rests on top of the block . The coefficient of static friction between the two blocks … 
math
A block of mass M1 = 2.7 kg rests on top of a second block of mass M2 = 4.7 kg, and the second block sits on a surface that is so slippery that the friction can be assumed to be zero (see the figure below). (a) If the coefficient of … 
Physics
A block with mass M = 5.00 kg rests on a frictionless table and is attached by a horizontal spring (k = 130 N/m) to a wall. A second block, of mass m = 1.25 kg rests on top of M. The blocks are displaced by 10 cm then released. If … 
Physics
A block M1 of mass 13.0 kg sits on top of a larger block M2 of mass 23.0 kg which sits on a flat surface. The kinetic friction coefficient between the upper and lower block is 0.440. The kinetic friction coefficient between the lower … 
physics
a block of mass M = 9.9 kg rests on a bracket of mass m = 5.5 kg. the bracket sits on a frictionless surface. the coefficients of friction between the block and the bracket on which it rests are .4 and .3. what is the maximum force …