Answer:
7.23407 [tex]\frac{m}{s^2}[/tex]
Explanation:
(I will not include units in calculations)
I'm assuming FBD's are already drawn, so I will work from there.
Let the 2.2kg block equal [tex]m_2[/tex], and the 20kg block equal [tex]m_1[/tex].
Summation equation for [tex]m_2[/tex]: [tex]\sum F_x=F_t_2-(F_f+F_g_x)=m_2a[/tex], [tex]\sum F_y=F_n-F_g_y=0[/tex]
Summation equation for [tex]m_1[/tex]: [tex]\sum F_y=F_g-F_t_1=m_1a[/tex]
Torque Summation Equation: [tex]\sum\tau=F_t_1*r-F_t_2*r=I\alpha[/tex]
Do some plugging in with the values given: [tex]\sum\tau=F_t_1*r-F_t_2*r=.5Mr^2\alpha[/tex]
Replace [tex]\alpha[/tex] with [tex]\frac{a}{r}[/tex], and cancel out the r's.
[tex]\sum\tau=F_t_1-F_t_2=.5Ma[/tex]
This step is important: Rearrange the force summation equation to solve for each tension force.
[tex]F_t_2=m_2a+F_f+F_g_x\\F_t_1=m_1g=m_1a[/tex]
Perform Substitution: [tex]\sum\tau=m_1g-m_1a-(m_2a+F_f+F_g_x)=.5Ma[/tex]
Now, we need to find the friction force and the horizontal component of the force of gravity.
Note that [tex]F_f=[/tex]μ[tex]F_n[/tex]
And based on our earlier summation equation: [tex]F_n=F_g_y[/tex]
First, break [tex]F_g[/tex] into x and y components. [tex]F_g_y=F_g\cos(\theta)[/tex], [tex]F_g_x=F_g\sin(\theta)[/tex]
Perform substitution with this and the fact that [tex]F_g=mg[/tex].
[tex]\sum\tau=m_1g-m_1a-(m_2a+\mu*m_2g\cos(\theta)+m_2g\sin(\theta))=.5Ma[/tex]
Solving for a, plugging in numbers yields an answer of 7.23407 [tex]\frac{m}{s^2}[/tex]
