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Question

A dozer pushes up a 100 kg spool of cable along a $20^ \circ$ incline road at a constant velocity as shown in the figure. The figure shows a dozer pushing a spool (diameter 400 mm) up a $20^ \circ$ incline. Point A is the contact point between the spool and the road, and Point B is the contact point between the dozer bucket and the spool. The coefficient of static friction between the dozer bucket and the spool (Point B) is 0.45, and coefficient of kinetic friction between road and the spool (Point A) is 0.15. 

Consider the spool only slides up the incline. The maximum normal force in N acting at Point B, is ___________ [rounded off to 1 decimal place]

1. Coordinate System and Assumptions

We define the \( x \)-axis parallel to the incline (pointing upwards) and the \( y \)-axis perpendicular to the incline. At constant velocity and with no rotation, the net force and net torque about the spool's center are zero.

2. Equilibrium Equations

Sum of forces in the \( y \)-direction (perpendicular to incline):

\[ N_A - f_B - mg \cos 20^\circ = 0 \implies N_A = mg \cos 20^\circ + f_B \]

Sum of forces in the \( x \)-direction (parallel to incline):

\[ N_B - f_A - mg \sin 20^\circ = 0 \implies N_B = mg \sin 20^\circ + f_A \]

Torque balance about the center ensures \( f_A = f_B \). Since it slides at A, \( f_A = \mu_k N_A = 0.15 N_A \).

3. Detailed Calculation

Substituting the friction relation into the \( y \)-balance equation:

\[ N_A = mg \cos 20^\circ + 0.15 N_A \] \[ 0.85 N_A = (100 \times 9.81) \times \cos 20^\circ \] \[ N_A \approx \frac{921.84}{0.85} = 1084.52 \text{ N} \]

Now, finding the friction force \( f_A \):

\[ f_A = 0.15 \times 1084.52 \approx 162.68 \text{ N} \]

Finally, calculating the normal force at B (\( N_B \)):

\[ N_B = mg \sin 20^\circ + f_A \] \[ N_B = (100 \times 9.81 \times \sin 20^\circ) + 162.68 \] \[ N_B \approx 335.52 + 162.68 = 498.2 \text{ N} \]
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