A 30m bar with a mass of 20kg is being lifted by a string. The bar makes an angle of 55 degrees above horizontal (between the bar and ground). The string is attached to the ceiling vertically.

What is the tension on the string?
What is the normal force given by the bar on the ground
what is the force of static friction given by the ground on the bar?

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the questions have to be answered. Physics is not a guessing game.

Are you sure that's 30 meters? That's over 1/2 and American football field long.

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To determine the tension on the string, we need to analyze the forces acting on the bar.

1. Tension on the string:
The gravitational force acting on the bar is given by the formula F = m * g, where m is the mass of the bar and g is the acceleration due to gravity (approximately 9.8 m/s^2). So, the gravitational force is F_gravity = 20 kg * 9.8 m/s^2 = 196 N.

Since the bar is at an angle of 55 degrees above horizontal, we need to resolve the gravitational force into its vertical and horizontal components. The vertical component of the gravitational force is given by F_vertical = F_gravity * sin(55 degrees) = 196 N * sin(55 degrees) ≈ 158.32 N.

The tension in the string is equal to the vertical component of the gravitational force because it counterbalances that force. So, the tension on the string is approximately 158.32 N.

2. Normal force on the bar from the ground:
The normal force acts perpendicular to the surface of contact. In this case, since the bar is not in contact with the ground, there is no normal force.

3. Force of static friction between the bar and the ground:
Since the bar is not in contact with the ground, there is no force of static friction.