A man drags a 71.5-kg crate across the floor at a constant velocity by pulling on a strap attached to the bottom of the crate. The crate is tilted 24.5° above the horizontal, and the strap is inclined 63.0° above the horizontal. The center of gravity of the crate coincides with its geometrical center, as indicated in the drawing. Find the magnitude of the tension in the strap.
To find the magnitude of the tension in the strap, we can use Newton's second law of motion.
First, let's break down the forces acting on the crate:
1. Weight (mg): The weight of the crate can be calculated by multiplying the mass (m) of the crate by the acceleration due to gravity (g). In this case, the weight acts vertically downwards.
2. Normal force (N): The normal force acts perpendicular to the surface of contact between the crate and the floor. Since the crate is at a constant velocity and not accelerating vertically, the normal force will be equal in magnitude and opposite in direction to the weight of the crate.
3. Tension force (T): The tension force in the strap is what allows the man to drag the crate. It acts along the direction of the strap, which is inclined at an angle of 63.0° above the horizontal.
4. Friction force (f): The friction force opposes the motion of the crate and acts parallel to the floor. Since the crate is moving at a constant velocity, the friction force will be equal in magnitude and opposite in direction to the tension force.
Now, let's resolve the forces along the horizontal and vertical axes:
Along the horizontal axis:
The tension force is the only force acting in this direction. We can resolve it into its horizontal and vertical components. The horizontal component of the tension force counterbalances the friction force.
Along the vertical axis:
The weight of the crate and the normal force cancel each other out vertically. The vertical component of the tension force counterbalances the weight of the crate.
Using trigonometry, we can calculate the horizontal and vertical components of the tension force:
Horizontal component (T_h):
T_h = T * cos(63.0°)
Vertical component (T_v):
T_v = T * sin(63.0°)
Since the crate is at a constant velocity, the sum of the forces along the horizontal axis is zero. Therefore, the horizontal component of the tension force (T_h) is equal to the friction force (f).
The friction force can be calculated by multiplying the coefficient of friction (μ) between the crate and the floor by the normal force (N). However, we first need to calculate the value of the normal force.
The angle at which the crate is tilted above the horizontal (24.5°) is the same as the angle between the vertical and the normal force. Therefore, we can use trigonometry to determine the value of the normal force:
N = mg * cos(24.5°)
Since the crate is at a constant velocity, the friction force (f) is equal to the horizontal component of the tension force (T_h):
f = T_h
Therefore, the normal force (N) can be expressed as:
f = μN
Solving the equation for N:
N = f / μ
Substituting the values, we can now calculate the magnitude of the tension in the strap:
T = sqrt(T_h^2 + T_v^2)