Question

A 20-cm long stick of m = 0.800 kg is lifted by a rope tied 6.0 cm from the upper end. The other end touches a smooth floor. The stick makes an angle of 0∘ with the floor. Find the magnitude of the normal reaction from the floor on the stick.

Answer 1

The magnitude of the normal reaction from the floor on the stick is approximately 18.27 N.

Step-by-step explanation:

To find the magnitude of the normal reaction from the floor on the stick, we can use the concept of torque. Torque is the product of the force applied on an object and the perpendicular distance from the point of rotation. In this case, the force is the weight of the stick, which can be calculated as mass times acceleration due to gravity (w = m * g). The perpendicular distance can be determined by subtracting the length of the stick above the point of rotation from the total length of the stick. Then we can calculate the torque by multiplying the force and the perpendicular distance (T = w * d). Since the stick is in equilibrium, the torque from the normal reaction is equal and opposite to the torque from the weight, so we can set them equal to each other and solve for the normal reaction (N): N * 6.0 cm = w * (20.0 cm - 6.0 cm).

• mass of the stick (m) = 0.800 kg
• length of the stick (L) = 20 cm
• distance from upper end to the point of rotation (d) = 6.0 cm
• acceleration due to gravity (g) = 9.8 m/s²

Using these values, we can calculate the magnitude of the normal reaction (N) as follows:

N * 0.06 m = (0.800 kg * 9.8 m/s²) * (0.14 m)

N = 0.800 kg * 9.8 m/s² * 0.14 m / 0.06 m

N ≈ 18.27 N