Question

Body A weighs 108 N, and body B weighs 35 N. The coefficients of friction between A and the incline are μs = 0.56 and μk = 0.25. Angle θ is 40°. Let the positive direction of an x axis be up the incline.
(a) Find the acceleration of A if A is initially at rest.
( ____ m/s2) ihat

(b) Find the acceleration of A if A is moving up the incline.
( ____ m/s2) ihat

(c) Find the acceleration of A if A is moving down the incline.
( ____ m/s2) ihat

Answer 1

weight of the body A , W_A = m_Ag = 108 N

mass of the block A, m_A = (108 N)/(9.8 m/s²)

                                      = 11.02 kg

weight of the body B , W_B =m_B g = 35 N

mass of the block B, m_A = (35 N)/(9.8 m/s²)

                                      = 3.57 kg

The coefficients of friction between A and the incline are

         μ_s = 0.56 and μ_k = 0.25.

angle θ = 40⁰

.......................................................................

a)

if the body A is initially at rest :

using Newton's second law of motion ,

T - f - m_Agsinθ = 0

and   m_Bg - T = 0

        T = m_Bg

hence ,

m_Bg - f - m_Agsinθ = 0

frictional force f = m_Bg - (m_Ag)sinθ

                         = (35 N) - (108 N)sin40

                         = -34.42 N

normal force N = (m_Ag) cosθ

                        = (108 N)coas40

                        = 82.73 N

so , maximum frictional force f_s,max   = μ_sN

                                                        = (0.56)(82.73 N)

                                                        = 46.33 N

from this ,

          f < f_s,max

so , the acceleration a = 0 m/s² i^

..........................................................................

b)

if A is moving up the incline :

using Newton's second law of motion ,

           m_Aa = T - f_k - m_Agsinθ    ............... (1)

    here , frictional force f_k = μ_kN

                                        = μ_km_Ag cosθ

equation (1) becomes

         m_Aa = T - μ_km_Ag cosθ - m_Agsinθ     ............. (2)  

for block B :

       m_Ba = m_Bg - T

tension T = m_Bg - m_Ba   ................ (3)

substitute the eq (3) in eq (2) , we get

       m_Aa = m_Bg - m_Ba - μ_km_Ag cosθ - m_Agsinθ

therefore , acceleration is

        a = [(m_Bg) - (m_Ag) sinθ - μ_k(m_Ag) cosθ] /(m_A + m_B)      ............. (4)   

substitute the given data in eq (4) , we get

       a = [(35 N) - (108 N) (sin40) - (0.25)(108 N) (cos40)] /( 11.02 kg+ 3.57 kg)

          = (-3.77 m/s²) i^ [objects are slowing down]

...............................................................

c)

if A is moving down the incline :

using Newton's second law of motion ,

           m_Aa = T + f_k - m_Agsinθ    ............... (5)

    here , frictional force f_k = μ_kN

                                        = μ_km_Ag cosθ

equation (5) becomes

         m_Aa = T + μ_km_Ag cosθ - m_Agsinθ     ............. (6)  

for block B :

       m_Ba = m_Bg - T

tension T = m_Bg - m_Ba   ................ (7)

substitute the eq (7) in eq (6) , we get

       m_Aa = m_Bg - m_Ba  + μ_km_Ag cosθ - m_Agsinθ

therefore , acceleration is

        a = [(m_Bg) - (m_Ag) sinθ + μ_k(m_Ag) cosθ] /(m_A + m_B)      ............. (8)   

substitute the given data in eq (4) , we get

       a = [(35 N) - (108 N) (sin40) + (0.25)(108 N) (cos40)] /( 11.02 kg+ 3.57 kg)

          = (-0.94 m/s²) i^ [objects are speeding up]