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PHYSICS HELP
PLEASE HELP ITS ABOUT ATWOOD MACHINES

PHYSICS HELP PLEASE HELP ITS ABOUT ATWOOD MACHINES-example-1

2 Answers

1 vote

Answer:

7.23407

Step-by-step explanation:

easy

User Sgt AJ
by
4.3k points
4 votes

Answer:

7.23407
(m)/(s^2)

Step-by-step 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
(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
(m)/(s^2)

User Holtavolt
by
4.5k points