Question

PHYSICS HELP PLEASE HELP ITS ABOUT ATWOOD MACHINES

Answer 1

Answer:

7.23407 $\frac{m}{s^2}$

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 $m_2$, and the 20kg block equal $m_1$.

Summation equation for $m_2$: $\sum F_x=F_t_2-(F_f+F_g_x)=m_2a$, $\sum F_y=F_n-F_g_y=0$

Summation equation for $m_1$: $\sum F_y=F_g-F_t_1=m_1a$

Torque Summation Equation: $\sum\tau=F_t_1*r-F_t_2*r=I\alpha$

Do some plugging in with the values given: $\sum\tau=F_t_1*r-F_t_2*r=.5Mr^2\alpha$

Replace $\alpha$ with $\frac{a}{r}$, and cancel out the r's.

$\sum\tau=F_t_1-F_t_2=.5Ma$

This step is important: Rearrange the force summation equation to solve for each tension force.

$F_t_2=m_2a+F_f+F_g_x\\F_t_1=m_1g=m_1a$

Perform Substitution: $\sum\tau=m_1g-m_1a-(m_2a+F_f+F_g_x)=.5Ma$

Now, we need to find the friction force and the horizontal component of the force of gravity.

Note that $F_f=$μ$F_n$

And based on our earlier summation equation: $F_n=F_g_y$

First, break $F_g$ into x and y components. $F_g_y=F_g\cos(\theta)$, $F_g_x=F_g\sin(\theta)$

Perform substitution with this and the fact that $F_g=mg$.

$\sum\tau=m_1g-m_1a-(m_2a+\mu*m_2g\cos(\theta)+m_2g\sin(\theta))=.5Ma$

Solving for a, plugging in numbers yields an answer of 7.23407 $\frac{m}{s^2}$