Q33E - University Physics with Modern Physics

Question

You are lowering two boxes, one on top of the other, down a ramp by pulling on a rope parallel to the surface of the ramp (Fig. E5.33). Both boxes move together at a constant speed of $15.0 \text{cm/s}$ . The coefficient of kinetic friction between the ramp and the lower box is $\mathbf{0}.\mathbf{444}$, and the coefficient of static friction between the two boxes is $\mathbf{0}.800$. (a) What force do you need to exert to accomplish this? (b) What are the magnitude and direction of the friction force on the upper box?

 

Figure E5.33

 

Step-by-Step Solution

Answer

1. The force need to exert to accomplish this is,  $57.8 \text{N}$
2. The magnitude of the force is $146.4 \text{N}$ and is in upward direction.

Step 1: Identification of the given data

 

The given data can be listed below as,

• The mass of upper box is, $m_2=32.0 \text{kg}$.
• The mass of lower box is, $m_1=48.0 \text{kg}$.
• The speed of the boxes is,$v=15.0 \text{cm/s}$ .
• The kinetic friction between the ramp and lower box is,${\mu }_k=0.444$ .
• The coefficient of static friction between the two boxes is,${\mu }_s=0.800$ .

Step 2: Significance of Coefficient of friction

The coefficient of friction can be defined as the ratio of frictional force to that of normal force when the two surfaces are in contact with each other.

Step 3: (a) Determination of force.

The expression for the balancing force on upper box can be expressed as,

  $F_{net}=m a$

Here m is the mass, is the acceleration.

 

Substitute 0 for a , in the above equation because boxes moves at constant speed.

 

$$\begin{gathered} F_{net}=m \left(0\right) \\ =0 \end{gathered}$$

Hence, the force is zero so the boxes are in equilibrium.

 

From the figure the expression for the angle can be expressed as,

$$\begin{gathered} \tan \theta =\frac{\text{perpendicular}}{\text{base}} \\ \tan \theta =\frac{2.50 \text{m}}{4.75 \text{m}} \\ \theta =27.8° \end{gathered}$$

Hence, the angle is, .$27.8°$

The expression for the net force along inclination can be expressed as,

 $F_{net}=F-m g \sin \theta +f_k$ 

Here g is the acceleration due to gravity, F is the force,  $f_k$is the force of friction.

 

Substitute 0 for $F_{net}$, ${\mu }_k m g \cos \theta$for $f_k$in the above equation.

$$\begin{gathered} 0=F-m g \sin \theta +{\mu }_k m g \cos \theta \\ F=m g \sin \theta -{\mu }_k m g \cos \theta \\ F=m g \left(\sin \theta -{\mu }_k\cos \theta \right) \end{gathered}$$

Here ${\mu }_k$is the coefficient of kinetic friction.

 

Substitute (32.0 kg + 48.0 kg) for m, $9.8 {\text{m/s}}^{\text{2}}$for g , and $27.8°$for$\theta$ , 0.444 for ${\mu }_k$in the above equation.

 $$\begin{gathered} F=(32.0 \text{kg} \text{ + 48.0} \text{kg)} \text{(9.8} {\text{m/s}}^2) \left(\sin (27.8°)-(0.444) \cos (27.8°)\right) \\ F=57.8 \text{N} \end{gathered}$$ 

Hence, required force is, 57.8 N.

 

Step 4: (b) Determination of magnitude and direction of the upper box

The expression for the balancing force on upper box can be expressed as,

 $F=m_2 g \sin \theta$

Substitute 32.0 kg for $m_2$,$9.8 {\text{m/s}}^{\text{2}}$for g, and $27.8°$for  $\theta$in the above equation.

 

$\begin{array}{rcl}F& =& (32.0 \text{kg}) (9.8 {\text{m/s}}^{\text{2}}) \sin (27.8°)\\ & =& 146.4 \text{N}\\ & & \end{array}$

Hence, required magnitude of the force is$146.4 \text{N}$ and is in upward direction along the inclination.