Question

You are lowering two boxes, one on top of the other, down the ramp shown in the figure (Figure 1) by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 18.0 cm/s . The coefficient of kinetic friction between the ramp and the lower box is 0.490, and the coefficient of static friction between the two boxes is 0.786.

What force do you need to exert to accomplish this?

What is the magnitude of the friction force on the upper box?

Answer 1

You are lowering two boxes, one on top of the other, down the ramp shown in the figure by pulling on a rope parallel to the surface of the ramp.

is 2.50 4.75m

Both boxes move together at a constant speed, v = 18 cm/s

a=0 a=0 ng

(a) Force is needed to exert to accomplish this which is given as :

According to an above free body diagram, we have

on vertical y-axis :     $\sum$F_y = m a_y

N_total = m_total g cos $\theta$                                                                

and   f_k = $\mu$_k N_total$\Rightarrow$$\mu$_k m_total g cos $\theta$    { eq.1 }

on horizontal x-axis :   $\sum$F_x = m a_x

f_k + T - m_total g sin $\theta$ = 0

T = m_total g sin $\theta$ - f_k       { eq.2 }

T = m_total g sin $\theta$ - $\mu$_k m_total g cos $\theta$

T = m_total g [sin $\theta$ - $\mu$_k cos $\theta$]                                                                             { eq.3 }

where, $\theta$ = angle = 27.7 degree

$\mu$_k = coefficient of kinetic friction between the ramp and the lower box = 0.49

inserting the values in eq.3

T = (80 kg) (9.8 ms/²) [sin 27.76⁰ - (0.49) cos 27.76⁰]

T = (784 N) [(0.4657) - (0.49) (0.8849)]

T = (784 N) (0.0321)

T = 25.1 N

(b) Magnitude of the friction force on the upper box which is given as :

$\sum$F_x = m a_x

f_s = m g sin $\theta$                                                                                            { eq.4 }

inserting the values in eq.4,

f_s = (32 kg) (9.8 m/s²) sin 27.76⁰

f_s = (313.6 N) (0.4657)

f_s = 146.1 N