Newton's second Law of motion. Coefficient of friction. Newton's first Law of motion. In Figure Q4 (b) the coefficient of static friction between the 3.00 kg crate and the 35.0° incline is 0.30.

The mass of the floodlight is $m = 20.0 \text{ kg}$. The acceleration due to gravity is $g = 9.81 \text{ m/s}^2$.

$$W = mg = (20.0 \text{ kg})(9.81 \text{ m/s}^2) = 196.2 \text{ N}$$

Taking torques about the hinge point (where the beam meets the pole), the sum of torques must be zero. Let $L$ be the length of the beam. The weight $W$ creates a clockwise torque, and the vertical component of the tension $T \sin(30.0^\circ)$ creates a counter-clockwise torque.

$$\sum \tau = 0$$ $$(T \sin(30.0^\circ))L - WL = 0$$ $$T \sin(30.0^\circ) = W$$

$$T = \frac{W}{\sin(30.0^\circ)} = \frac{196.2 \text{ N}}{0.5} = 392.4 \text{ N}$$

Rounding to three significant figures:

$$T = \boxed{\text{392 N}}$$

The sum of horizontal forces must be zero. The horizontal component of tension $T \cos(30.0^\circ)$ acts to the left. The horizontal reaction force from the pole $R_x$ acts to the right.

$$\sum F_x = 0$$ $$R_x - T \cos(30.0^\circ) = 0$$ $$R_x = T \cos(30.0^\circ) = (392.4 \text{ N}) \cos(30.0^\circ)$$ $$R_x = (392.4 \text{ N})(0.8660) = 339.82 \text{ N}$$

Rounding to three significant figures:

$$R_x = \boxed{\text{340 N (to the right)}}$$

The sum of vertical forces must be zero. The vertical component of tension $T \sin(30.0^\circ)$ acts upwards. The weight $W$ acts downwards. The vertical reaction force from the pole $R_y$ acts upwards.

$$\sum F_y = 0$$ $$R_y + T \sin(30.0^\circ) - W = 0$$

From Step 2, we know $T \sin(30.0^\circ) = W$.

$$R_y + W - W = 0$$ $$R_y = \boxed{\text{0 N}}$$

The vertical component of the force exerted by the pole on the beam is zero.