The block on this incline weighs 100 kg and is connected by a cable and pulley to a weight of 10 kg. If the coefficient of friction between

Question

The block on this incline weighs 100 kg and is connected by a cable and pulley to a weight of 10 kg. If the coefficient of friction between the block and incline is o.3, the block will:

in progress 0
Đan Thu 5 years 2021-08-26T03:28:02+00:00 1 Answers 41 views 0

Answers ( )

  1. MichaelMet
    0
    2021-08-26T03:29:47+00:00
    Reply

    Answer:

    a. 94.54 N

    b. 0.356 m/s^2

    Explanation:

    Given:-

    – The mass of the inclined block, M = 100 kg

    – The mass of the vertically hanging block, m = 10 kg

    – The angle of inclination, θ = 20°

    – The coefficient of friction of inclined surface, u = 0.3

    Find:-

    a) The magnitude of tension in the cable

    b) The acceleration of the system

    Solution:-

    – We will first draw a free body diagram for both the blocks. The vertically hanging block of mass m = 10 kg tends to move “upward” when the system is released.

    – The block experiences a tension force ( T ) in the upward direction due the attached cable. The tension in the cable is combated with the weight of the vertically hanging block.

    – We will employ the use of Newton’s second law of motion to express the dynamics of the vertically hanging block as follows:

                            T - m*g = m*a\\\\  … Eq 1

    Where,

                  a: The acceleration of the system

    – Similarly, we will construct a free body diagram for the inclined block of mass M = 100 kg. The Tension ( T ) pulls onto the block; however, the weight of the block is greater and tends down the slope.

    – As the block moves down the slope it experiences frictional force ( F ) that acts up the slope due to the contact force ( N ) between the block and the plane.

    – We will employ the static equilibrium of the inclined block in the normal direction and we have:

                            N - M*g*cos ( Q )= 0\\\\N = M*g*cos ( Q )

    – The frictional force ( F ) is proportional to the contact force ( N ) as follows:

                            F = u*N\\\\F = u*M*g *cos ( Q )

    – Now we will apply the Newton’s second law of motion parallel to the plane as follows:

                           M*g*sin(Q) - T - F = M*a\\\\M*g*sin(Q) - T -u*M*g*cos(Q) = M*a\\ .. Eq2

    – Add the two equation, Eq 1 and Eq 2:

                          M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g = a* ( M + m )\\\\a = \frac{M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g}{M + m} \\\\a = \frac{100*9.81*sin ( 20 ) - 0.3*100*9.81*cos ( 20 ) - 10*9.81}{100 + 10}\\\\a = \frac{-39.12977}{110} = -0.35572 \frac{m}{s^2}

    – The inclined block moves up ( the acceleration is in the opposite direction than assumed ).

    – Using equation 1, we determine the tension ( T ) in the cable as follows:

                         T = m* ( a + g )\\\\T = 10*( -0.35572 + 9.81 )\\\\T = 94.54 N

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )