Realizing that she often doesn't have her students' full attention during class, a professor devises an elaborate device on which to stand while she lectures. The pulley is placed so that the string makes a 45.0-degree angle with the beam. The beam is uniform, 5.00 meters long, and has weight . The professor stands 2.00 meters from the pivot point and has weight .
If the professor's mass is 60 kilograms and the mass of the beam is 20 kilograms, what is the magnitude of the normal force exerted by the beam on the professor? Use 10 meters per second squared for the magnitude of the acceleration due to gravity.
Express your answer in newtons to two significant figures.
To find the magnitude of the normal force exerted by the beam on the professor, we can start by analyzing the forces acting on the system.
1. Weight of the professor: The weight of the professor can be calculated using the formula:
weight_professor = mass_professor * acceleration_due_to_gravity
weight_professor = 60 kg * 10 m/s^2 = 600 N
2. Weight of the beam: The weight of the beam can be calculated similarly:
weight_beam = mass_beam * acceleration_due_to_gravity
weight_beam = 20 kg * 10 m/s^2 = 200 N
3. Tension in the string: The tension in the string can be found by resolving the weight of the beam into its horizontal and vertical components. Since the string makes a 45.0-degree angle with the beam, the vertical component of the weight is equal to the tension in the string. Using trigonometry, we can calculate this component:
vertical_component = weight_beam * sin(45.0 degrees)
vertical_component = 200 N * sin(45.0) ≈ 141.42 N
4. Sum of the vertical forces: The sum of the vertical forces must be zero for the beam to be in equilibrium. The normal force exerted by the beam on the professor cancels out a part of the professor's weight, so it can be calculated as:
normal_force = weight_professor - vertical_component
normal_force = 600 N - 141.42 N ≈ 458.58 N
Therefore, the magnitude of the normal force exerted by the beam on the professor is approximately 458.58 N.
To find the magnitude of the normal force exerted by the beam on the professor, we can start by analyzing the forces acting on the beam.
The forces acting on the beam include:
1. The weight of the beam, acting downward vertically.
2. The tension in the string, acting at an angle of 45.0 degrees with the beam.
3. The normal force exerted by the professor, acting perpendicular to the beam.
Since the beam is in equilibrium (not rotating or accelerating), the sum of the forces in both the horizontal and vertical directions must be zero.
Considering the vertical forces:
1. The weight of the beam acting downward is given by the formula: weight = mass x acceleration due to gravity. So, the weight of the beam is (20 kg) x (10 m/s^2) = 200 N.
2. The vertical component of the tension in the string can be found using trigonometry. The angle between the string and the vertical direction is 45.0 degrees. The vertical component of the tension force is (tension) x (cosine of 45.0 degrees).
Now, considering the horizontal forces:
3. The horizontal component of the tension in the string can also be found using trigonometry. The angle between the string and the horizontal direction is 45.0 degrees. The horizontal component of the tension force is (tension) x (sine of 45.0 degrees).
4. There are no other horizontal forces acting on the beam, so the horizontal forces sum up to zero.
Since there is no rotation or acceleration, the magnitudes of the vertical forces should be equal. Therefore, the magnitude of the normal force exerted by the beam on the professor is equal to the sum of the weight of the beam and the vertical component of the tension in the string.
To find the vertical component of the tension in the string, we use the following equation:
Vertical component of tension = (tension) x (cosine of 45.0 degrees).
Substituting known values:
Vertical component of tension = (tension) x (cos(45.0 degrees)).
Next, we can set up an equation using Newton's Second Law (F = ma) for the vertical forces:
Normal force - weight of the beam - vertical component of tension = 0.
Plugging in the given values:
Normal force - 200 N - [(tension) x (cos(45.0 degrees))] = 0.
Finally, solve this equation to find the magnitude of the normal force exerted by the beam on the professor.