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PtichkaEL [24]
3 years ago
6

Block B in rests on a surface for which the static and kinetic coefficients of friction are 0.59 and 0.40, respectively. The rop

es are massless.What is the maximum mass of block A for which the system remains in static equilibrium?

Physics
1 answer:
Pavlova-9 [17]3 years ago
4 0

m_A = 11.8\:\text{kg}

Explanation:

Let T_A\:\text{and}\:T_B be the tensions on ropes on blocks A and B, respectively, T_C be the tension at an angle and f_s be the frictional force. Adopt the usual sign convention (right and up are positive and left and down are negative). To achieve static equilibrium, we demand that the net forces acting on the objects to be zero. Now let's apply Newton's 2nd law (NSL) to the objects.

Block A:

y:\:\:\:T_A - m_Ag = 0 (1)

Block B:

x:\:\:\:T_B - f_s = T_B - \mu_sN = 0 (2)

y:\:\:\:N - m_Bg = 0 (3)

<u>Intersection</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>ropes</u><u>:</u>

x:\:\:\:T_C\cos{45} - T_B = 0 \Rightarrow T_B = T_C\cos{45} (4)

y:\:\:\:T_C\sin{45} - T_A = 0 \Rightarrow T_A = T_C\sin{45} (5)

Using Eq(4) and Eq(3) on Eq(2) and Eq(5) on Eq(1), we get

T_C\cos{45} = \mu m_Bg (6)

T_C\sin{45} = m_Ag (7)

Dividing Eq(7) by Eq(6), we get

\tan{45} = \dfrac{m_A}{\mu m_B} = 1

Solving for m_A, we find that the mass needed to keep the blocks steady and motionless is

m_A = \mu m_B = (0.59)(20\:\text{kg})

\:\:\:\:\:\:\:= 11.8\:\text{kg}

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