Question

block 1 of mass m1 is placed on block 2 of mass m2 which is then placed on a table. a string connecting block 2 to a hanging mass m passes over a pulley attached to one end of the table, as shown above. the mass and friction of the pulley are negligible. the coefficients of friction between blocks 1 and 2 and between block 2 and the tabletop are nonzero and are given in the following table.

	coefficient between blocks 1 and 2	coefficient between block 2 and the tabletop
kinetic	$mu_{k1}$	$mu_{k2}$

express your answers in terms of the masses, coefficients of friction, and g, the acceleration due to gravity.
a. suppose that the value of m is small enough that the blocks remain at rest when released. for each of the following forces, determine the magnitude of the force and draw a vector on the block provided to indicate the direction of the force if it is nonzero.
i. the normal force n1 exerted on block 1 by block 2
ii. the friction force f1 exerted on block 1 by block 2
iii. the force t exerted on block 2 by the string m2
iv. the normal force n2 exerted on block 2 by the tabletop m2
v. the friction force f2 exerted on block 2 by the tabletop
b. determine the largest value of m for which the blocks can remain at rest.
c. now suppose that m is large enough that the hanging block descends when the blocks are released. assume that blocks 1 and 2 are moving as a unit (no slippage). determine the magnitude of their acceleration.
d. now suppose that m is large enough that as the hanging block descends, block 1 is slipping on block 2. determine each of the following.
i. the magnitude a1 of the acceleration of block 1
ii. the magnitude a2 of the acceleration of block 2

Explanation:

Step1: Analyze normal force on block 1

In the vertical - direction for block 1, the net - force is zero since it is at rest. The normal force $N_1$ exerted on block 1 by block 2 balances the weight of block 1. So, $N_1 = m_1g$.

Step2: Analyze friction force on block 1

Since the blocks are at rest and there is no relative motion or tendency of relative motion between block 1 and block 2 in the horizontal direction, the friction force $f_1$ exerted on block 1 by block 2 is $f_1 = 0$.

Step3: Analyze tension force on block 2

For the system to be at rest, the tension $T$ in the string is equal to the weight of the hanging mass $M$. So, $T = Mg$.

Step4: Analyze normal force on block 2

In the vertical - direction for block 2, the normal force $N_2$ exerted on block 2 by the tabletop balances the combined weight of block 1 and block 2. So, $N_2=(m_1 + m_2)g$.

Step5: Analyze friction force on block 2

The maximum static friction force $f_2$ that can act on block 2 is given by the static - friction formula. Since the blocks are at rest, the friction force $f_2$ balances the tension force $T$. The maximum static friction force $f_2=\mu_{s2}N_2=\mu_{s2}(m_1 + m_2)g$. And $f_2 = T$ for the blocks to remain at rest, so $f_2=\mu_{s2}(m_1 + m_2)g$.

Step6: Find the maximum value of $M$ for the blocks to remain at rest

At the verge of moving, the tension $T$ (which is $T = Mg$) is equal to the maximum static - friction force on block 2. So, $Mg=\mu_{s2}(m_1 + m_2)g$, and $M=\mu_{s2}(m_1 + m_2)$.

Step7: Analyze the acceleration when the blocks move as a unit

For the combined system of blocks 1 and 2 and the hanging mass $M$, using Newton's second law $F = ma$. The net - force on the system is $F_{net}=Mg-\mu_{k2}(m_1 + m_2)g$. The total mass of the system is $M + m_1 + m_2$. So, $a=\frac{Mg-\mu_{k2}(m_1 + m_2)g}{M + m_1 + m_2}$.

Step8: Analyze the acceleration of block 1 when it slips

When block 1 slips on block 2, the friction force acting on block 1 is the kinetic - friction force. Using Newton's second law for block 1, $f_{k1}=\mu_{k1}N_1=\mu_{k1}m_1g$. And $f_{k1}=m_1a_1$, so $a_1=\mu_{k1}g$.

Step9: Analyze the acceleration of block 2 when block 1 slips

For block 2, using Newton's second law $T - f_{k1}-f_{k2}=m_2a_2$. The tension $T = Mg$, $f_{k1}=\mu_{k1}m_1g$, and $f_{k2}=\mu_{k2}(m_1 + m_2)g$. So, $Mg-\mu_{k1}m_1g-\mu_{k2}(m_1 + m_2)g=m_2a_2$, and $a_2=\frac{Mg-\mu_{k1}m_1g-\mu_{k2}(m_1 + m_2)g}{m_2}$.

Answer:

a.
i. $N_1 = m_1g$
ii. $f_1 = 0$
iii. $T = Mg$
iv. $N_2=(m_1 + m_2)g$
v. $f_2=\mu_{s2}(m_1 + m_2)g$
b. $M=\mu_{s2}(m_1 + m_2)$
c. $a=\frac{Mg-\mu_{k2}(m_1 + m_2)g}{M + m_1 + m_2}$
d.
i. $a_1=\mu_{k1}g$
ii. $a_2=\frac{Mg-\mu_{k1}m_1g-\mu_{k2}(m_1 + m_2)g}{m_2}$