Two blocks of mass m1 = 11 kg and m2 = 13 kg are connected by a string (that doesn’t stretch) that passes over a frictionless pulley as shown in the diagram (attached). The side that m1 is on has a coefficient of static friction µs = .79 and a coefficient of kinetic friction µk = .12 while the side that m2 is on is frictionless and θ = 45°. A. Assuming the block of mass m2 is accelerating down its side, draw the free body diagrams for each block. B. Using Newton’s laws, set up but DO NOT SOLVE the equations necessary to find the acceleration of the system. C. If the string is cut at the pulley, find the subsequent accelerations of the two blocks. Having a lot of trouble with this physics homework question.

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The central task in such problems is to draw force diagrams correctly. After that they become not so difficult.

A . If `m_2` is accelerating down, then `m_1` is accelerating up. Their accelerations `a_1` and `a_2` (vectors) have equal magnitude `a.` The same we can say about the traction...

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The central task in such problems is to draw force diagrams correctly. After that they become not so difficult.

A. If `m_2` is accelerating down, then `m_1` is accelerating up. Their accelerations `a_1` and `a_2` (vectors) have equal magnitude `a.` The same we can say about the traction forces `F_(T1)` and `F_(T2)` and their magnitude `F_T` (from the string).

Also, if `m_1` is moving, its friction is the kinetic one.

Please look at the picture. `N_1` and `N_2` are the reaction forces, `F_(f1)` is the friction force. The friction force acts to the opposite direction to the movement (down along its side for `m_1`). It is zero for `m_2` (given).

B. Newton's Second law says that

`m_1a_1=m_1g+F_T1+N_1+F_(f1)` and

`m_2a_1=m_2g+F_T2+N_2.`

All summands are vectors here.

Also we know that

`F_T=|F_(T1)|=|F_(T2)|,` `a=|a_1|=|a_2|,` `|F_(f1)|=mu_k*|N_1|.`

To solve these equations we have to project them on some axes. The most convenient are along the side and perpendicular to that side.

 

C. If the string is cut. The forces `F_(T1)` and `F_(T2)` vanish and Newton's Second law equations become

`m_1a_1=m_1g+N_1+F_(f1)`  and

`m_2a_2=m_2g+N_2.`

Note that `F_(f1)` is upside the hill now (`m_1` "wants" down).

For `m_2` project its equation to the axis down along its side and obtain

`m_2a_2=m_2gsin(theta),` or `a_2=g*sin(theta) approx 6.9 (m/s^2).`

 

For `m_1` two projections are needed. For the downward axis along its side

`m_1a_1=m_1gsin(theta)-F_(f1),` so `a_1=gsin(theta)-F_(f1)/m_1.`

`g sin(theta) approx 6.9 m/s^2.`

For the perpendicular axis `N_1=m_1gcos(theta).`

a) If `m_1` will move down, friction will be kinetic and `F_(f1)=mu_k*N_1=mu_k m_1 g cos(theta),` so

`F_(f1)/m_1=mu_k g cos(theta) approx 0.8 (m/s^2)`

and `a_1 approx 6.9-0.8=6.1 (m/s^2).`

b) if `m_1` will be stopped manually AND released, the maximum static friction will be `m_1*mu_s*g cos(theta) approx m_1g*0.56,` which is less than `m_1g*sin(theta) approx m_1g*0.71`. So `m_1` couldn't be in rest and the a) variant is the only possible.

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