Question

In the figure below, A is a 44 N block and B is a 22 N block. The coefficient of static friction between A and the table is 0.20 and the minimum weight of the block C to make Block A and C static is 66 N (a) Block C suddenly is lifted off A. Calculate the acceleration of block A, if the coefficient of kinetic friction between A and the table is 0.15.

Answer 1

2.29 m/s²

Step-by-step explanation

The free body diagram of this situation, when block C is lifted off A is

Now we know that blocks A and B are moving, so the equations according with Newton's second law of motion for each block are:

Block A:

`\begin{gathered} F_T-F_f=m_A\cdot a \\ F_N-F_(gA)=0 \end{gathered}`

Block B (there are not horizontal forces):

`F_T-F_(gB)=-m_B\cdot a`

From the equation for block B, we have that the tension of the rope is:

`\begin{gathered} F_T=F_(gB)-m_B\cdot a \\ F_T=m_B\cdot g-m_B\cdot a \\ F_T=m_B(g-a) \end{gathered}`

And mB is:

`m_B=(F_(gB))/(g)`

So the tension force is:

`F_T=(F_(gB))/(g)(g-a)`

The kinetic friction force, when the coefficient of kinetic friction betwen the surfaces is μk is:

`F_f=\mu_k\cdot F_N`

The normal force, from the second equation of block A is:

`F_N=F_(gA)=44N`

So the friction force between block A and the table is:

`F_f=0.15\cdot44N=6.6N`

Now, using the first equation for block A, we can find its acceleration. But before doing that, we have to find the mass of block A - remember that we have the weight, not the mass. This we find with the equation of weight:

`\begin{gathered} F_(gA)=m_A\cdot g \\ m_A=(F_(gA))/(g) \end{gathered}`

Replacing Ft, Ff and mA into the first equation of block A:

`F_T-F_f=m_A\cdot a`
`(F_(gB))/(g)(g-a)-6.6N=(F_(gA))/(g)\cdot a`

We have an equation which we can solve for a. Apply distributive property in the first term:

`\begin{gathered} (F_(gB)\cdot g)/(g)-(F_(gB)\cdot a)/(g)-6.6N=(F_(gA))/(g)\cdot a \\ F_(gB)-(F_(gB)\cdot a)/(g)-6.6N=(F_(gA))/(g)\cdot a \end{gathered}`

Then add the second term from both sides:

`\begin{gathered} F_(gB)-(F_(gB)\cdot a)/(g)+(F_(gB)\cdot a)/(g)-6.6N=(F_(gA))/(g)\cdot a+(F_(gB)\cdot a)/(g) \\ F_(gB)-6.6N=(F_(gA))/(g)\cdot a+(F_(gB))/(g)\cdot a \end{gathered}`

Take a as a common factor from the right side of the equation:

`F_(gB)-6.6N=((F_(gA))/(g)+(F_(gB))/(g))a`

And then divide by the coefficient of a:

`a=(F_(gB)-6.6N)/(((F_(gA))/(g)+(F_(gB))/(g)))`

Since g is a common denominator we can simplify the expression a little before replacing with values:

`a=\frac{F_(gB)-6.6N_{}}{(F_(gA)+F_(gB))/(g)}=(g(F_(gB)-6.6N))/(F_(gA)+F_(gB))`

Now we can replace and find a:

`a=(9.81m/s^2(22N-6.6N))/(44N+22N)=2.29m/s^2`