In the arrangement the coefficient of friction between the 2 blocks is#mu=1//2#What is the force of friction acting between the 2 blocks?

Question

Charlotte Aaron · Accepted Answer

#8N#

For ,the two blocks to move together,we can say the frictional force acting at the interface of the two blocks must supply the required force(#ma=2a#) for the smaller block to move forward.

So,we can say for the larger block, #F-f=4a# (where,#a# is the net acceleration)

or, #f=F-4a=24-4a#

And,for the smaller block, #f=2a#

So,comparing both, #24-4a=2a#

or, #a=4 ms^-2#

So,#f=8N#

Let's,see maximum value of frictional force that can act at their interface ,it is #mumg=1/2*2*9.8=9.8N#

So,required frictional force is enough for the two blocks to move together.

Note,frictional force will not act by #9.8 N#,as the #2*a# amount of force is acting backwards relative to the block(thinking from non inertial frame of reference),so frictional force will act just by that amount,so that it will have no relative motion backward,i.e will go along with the lower block,if value of #ma# would have exceeded maximum value of frictional force,then the smaller block would have started moving backwards due to that extra amount of force.

Nora Arzate · Answer

#8# [N]

#m_1 = 2# #m_2 = 4# #mu = 1/2# #g = 10# Supposing no slipping the set is accelerated as #(m_1+m_2) alpha_2 = F# or #alpha_2 = F/(m_1+m_2)# During carrying the upper block is submitted to a traction force of #m_1 alpha_2# this force compared with the maximum static traction force gives #mu m_1 g < m_1 alpha_2 = m_1/(m_1+m_2) F# so no slip occurs and the upper block is actuated with a force of # m_1/(m_1+m_2) F = 8#[N]

Charlotte Aaron · Answer

The force of friction acting on the 2 kg block will be 8 N in the direction of the 24 N force.

Analysis to determine whether or not the 2 kg block slips: If slipping does not occur, what will be the acceleration of the blocks? #F = 24 N = m*a = 6 kg*a#, therefore #a = (24 N)/(6 kg) = 4 m/s^2# Would the friction be able to support the 2 kg block accelerating at #4 m/s^2#? The 2 kg block would need a force, #F_2 = 2 kg*4 m/s^2 = 8 N# to accelerate at #4 m/s^2#. Will the force of friction, #F_f#, be able to provide 8 N? #F_f = mu*N = mu*m*g# #F_f = 1/2*2 kg*10 m/s^2 = 10 N# Yes, #F_f# can, and will, provide the 8 N required for the 2 kg block to accelerate at #4 m/s^2#. So, the 2 kg block will not slip. I hope this helps, Steve

Carson Alexander · Answer

undefined The force of friction acting between the two blocks can be calculated using the formula for frictional force, which is given by $ F_f = \mu \cdot N $, where $ \mu $ is the coefficient of friction and $ N $ is the normal force between the surfaces. If the normal force is known, we can simply multiply it by the coefficient of friction to find the frictional force. However, since the normal force depends on the weight of the object and the surface it's resting on, it needs to be calculated first before finding the frictional force. If the masses of the blocks are known, we can use the equation $ N = mg $, where $ m $ is the mass of the block and $ g $ is the acceleration due to gravity (approximately $ 9.8 \, \text{m/s}^2 $). Once the normal force is determined, we can then use it in the frictional force formula to find the force of friction between the two blocks.