A sledge loaded with bricks has a total mass of 18.9 kg and is pulled at constant speed by a rope inclined at 19.3° above the horizontal. The sledge moves a distance of 19.8 m on a horizontal surface. The coefficient of kinetic friction between the sledge and surface is 0.500. What is the tension on the rope?
The horizontal component of the tension in the rope, T cos 19.3, will equal the friction force, since the sledge is not accelerating.
Compute the friction force using the kinetic friction coefficient, making sure that you take into account the vertical component of the rope force, which reduces the normal and friction forces.
Then solve for T
To find the tension on the rope, we need to first calculate the force of kinetic friction acting on the sledge. We can use the formula for kinetic friction:
\( f_k = \mu_k \cdot N \)
where \( f_k \) is the force of kinetic friction, \( \mu_k \) is the coefficient of kinetic friction, and \( N \) is the normal force.
The normal force is the force exerted by the surface perpendicular to it. In this case, the sledge is on a horizontal surface, so the normal force is equal in magnitude and opposite in direction to the vertical force due to gravity:
\( N = mg \)
where \( m \) is the mass of the sledge and \( g \) is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the given values, we have:
\( N = 18.9 \, \text{kg} \times 9.8 \, \text{m/s}^2 \)
Next, we can calculate the force of kinetic friction:
\( f_k = 0.500 \times 18.9 \, \text{kg} \times 9.8 \, \text{m/s}^2 \)
Now, we need to find the force along the incline (parallel to the inclined plane). This force is equal in magnitude to the tension in the rope:
\( T = f_k + mg \sin(\theta) \)
where \( T \) is the tension, \( \theta \) is the angle of inclination (19.3°), and \( \sin(\theta) \) represents the component of gravity along the incline.
Finally, we can calculate the tension on the rope:
\( T = 0.500 \times 18.9 \, \text{kg} \times 9.8 \, \text{m/s}^2 + 18.9 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \sin(19.3°) \)
By evaluating this expression, we can find the numerical value of the tension on the rope.