Question

a 50.0-pound box is pulled by a rope across a level surface at constant velocity. the rope has an angle of 30.0 degrees above the horizontal, and the tension in the rope is 150 n. what is the magnitude of the normal force from the floor on the box? multiple choice 106 n 148 n 1184 n 1078 n

Explanation:

Step1: Convert weight to Newtons

The weight of the box is 50.0 pounds. We know that \( 1\ \text{pound} = 4.44822\ \text{N} \), so the weight \( W \) in Newtons is \( W = 50.0\times4.44822 = 222.411\ \text{N} \).

Step2: Analyze vertical forces

In the vertical direction, the forces are the weight of the box (\( W \)) downward, the vertical component of the tension (\( T_y \)) upward, and the normal force (\( N \)) upward. Since the box is moving at constant velocity, the net vertical force is zero. So \( N + T_y = W \), which means \( N = W - T_y \).
The vertical component of the tension is \( T_y = T\sin\theta \), where \( T = 150\ \text{N} \) and \( \theta = 30.0^\circ \). So \( T_y = 150\times\sin(30.0^\circ)=150\times0.5 = 75\ \text{N} \).

Step3: Calculate normal force

Substitute \( W = 222.411\ \text{N} \) and \( T_y = 75\ \text{N} \) into the equation for \( N \): \( N = 222.411 - 75 = 147.411\approx148\ \text{N} \).

Answer:

148 N