Understanding the Problem
To find the tension required to pull a crate at a constant velocity, we must apply Newton's First Law. Since the velocity is constant, the net force acting on the crate must be zero ($F_{net} = 0$).
Forces Involved
- Tension ($T$): The force exerted by the rope, acting at an angle of $30^\circ$ above the horizontal.
- Weight ($W$): Acting downward, $W = 500\text{ N}$.
- Normal Force ($N$): Acting upward from the surface.
- Kinetic Friction ($f_k$): Acting against the direction of motion, $f_k = \mu_k N$.
Step-by-Step Solution
1. Resolve forces into components:
- Horizontal: $T_x = T \cos(30^\circ)$
- Vertical: $T_y = T \sin(30^\circ)$
2. Equilibrium in the Vertical Direction (y-axis): $$N + T_y = W$$ $$N = W - T \sin(30^\circ) = 500 - T(0.5)$$
3. Equilibrium in the Horizontal Direction (x-axis): Since velocity is constant, the horizontal force equals the frictional force: $$T_x = f_k$$ $$T \cos(30^\circ) = \mu_k N$$
4. Substitute $N$ into the horizontal equation: $$T \cos(30^\circ) = 0.40(500 - 0.5T)$$ $$T(0.866) = 200 - 0.2T$$ $$T(0.866 + 0.2) = 200$$ $$T(1.066) = 200$$ $$T = \frac{200}{1.066} \approx 187.6\text{ N}$$
Intuition
When you pull at an angle, part of your force helps lift the crate, which reduces the normal force $N$. Because $f_k$ depends directly on $N$, pulling upward actually reduces the friction opposing your motion compared to pulling perfectly horizontally.