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Homework Statement
Two blocks A and B are connected by an ideal rope R, both blocks lie on a horizontal surface S. The coefficient of static friction ([tex]\mu_s = 0.30[/tex]) and kinetic friction ([tex]\mu_k = 0.20[/tex]) between the blocks and the surface S are the same for both blocks. The mass of block A is [tex]m_A = 6.12 kg[/tex]. The mass of block B is [tex]m_B = 7.51 kg[/tex]. A horizontal force [tex]F_1 = 50.39 N[/tex] pushes the block A, a horizontal force [tex]F_2 = 26.81 N[/tex] pushes the block B.
Assume [tex]g = 9.8 m/s^2[/tex].
(See picture)
What is the acceleration of the block and the tension on the rope?
Homework Equations
[tex]F = ma[/tex]
[tex]f_k = \mu_k F_n[/tex]
[tex]f_s(max) = \mu_s F_n[/tex]
The Attempt at a Solution
Since [tex]m_A = 6.12 kg[/tex], then [tex]F_n(A) = 6.12 \times 9.8 = 59.98 N[/tex]
Since [tex]m_B = 7.51 kg[/tex], then [tex]F_n(B) = 7.51 \times 9.8 = 73.60 N[/tex]
[tex]F_1[/tex] is greater than [tex]f_s(max)[/tex], therefore the system is in motion and the friction is kinetic.
[tex]f_k(A) = 0.2 \times 59.98 = 12.00 N[/tex]
[tex]f_k(B) = 0.2 \times 73.60 = 14.72 N[/tex]
Let the direction of [tex]F_1[/tex] be the positive one, and T be the tension magnitude.
So by Newton's Second Law, we have:
[tex]F_1  f_k(A)  T = m_A \times a[/tex]
[tex]T  F_2  f_k(B) = m_B \times a[/tex]
Adding the two equations above yields:
[tex]F_1  F_2  f_k(A)  f_k(B) = (m_A + m_B) \times a[/tex]
[tex]a = \frac{F_1  F_2  f_k(A)  f_k(B)}{m_A + m_B}[/tex]
[tex]a = \frac{50.39  26.81  12.00  14.72}{6.12 + 7.51} = 0.23 m/s^2[/tex]
But why this negative acceleration??? Clearly my answer is wrong, since I assumed the positive direction the direction of [tex]F_1[/tex], which is greater than [tex]F_2[/tex]. Please, help me.
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