A 3.0 m long rigid beam with a mass of 100 kg is supported at each end. An 60 kg student stands 2.0 m from support 1. How much upward force does each support exert on the beam?

support 1-
support 2-

All of that stuff but than multiple R1, R2 by 9.8

.......100..60......

..........v......v.......
----------------------
^.......................^
R1.................R2

Take moments about R1,
R2*3=100*1.5+60*2.0
Take moments about R2
R1*3=100*1.5+60*1.0
Solve for R1 and R2. They should add up to 100+60.

Well, well, well, looks like we have some physics here! Let's see if I can clown around with these numbers for you.

To find out how much upward force each support exerts on the beam, we need to consider the forces acting on it. There's the gravitational force acting downwards due to the mass of the beam and the student, and there's also the upward force from each support.

Now, to take the weight off your shoulders, I have a formula that might help. The sum of the forces acting on the beam in the vertical direction should equal zero (since the beam is not moving up or down). So, let's do some calculations!

The gravitational force acting on the beam is given by mass times gravity, which is (100 kg + 60 kg) * 9.8 m/s². That should give you the total downward force.

Now, the distance the student is standing from support 1 doesn't directly affect the upward force at support 2. We only need it to find the weight of the student.

To divide the total downward force evenly between the supports, we just divide it by 2 because each support is sharing the load.

So, the upward force at each support is half of the total downward force. But hey, don't expect me to do all the math! Give it a shot, and let's see if you can calculate those forces.

To determine the upward force exerted by each support on the beam, we need to analyze the forces acting on the beam in equilibrium. The sum of all the vertical forces must be zero since the beam is not accelerating vertically.

Let's start with support 1. We can draw a free-body diagram of the forces acting on the beam at support 1:

1. The weight of the beam acts downward from the center of gravity of the beam, which is at its midpoint. The weight can be calculated as the mass of the beam multiplied by the acceleration due to gravity: weight1 = 100 kg * 9.8 m/s^2.

2. The reaction force at support 1 acts upward to balance the weight of the beam and the force applied by the student. Let's call this force R1.

3. The force applied by the student acts downward at a distance of 2.0 m from support 1. This force can be calculated as the mass of the student multiplied by the acceleration due to gravity: force_student = 60 kg * 9.8 m/s^2.

Since the beam is in equilibrium, the sum of the vertical forces at support 1 must be zero:

R1 - weight1 - force_student = 0.

Substituting the known values, we have:

R1 - (100 kg * 9.8 m/s^2) - (60 kg * 9.8 m/s^2) = 0.

Now, let's solve for R1:

R1 = (100 kg * 9.8 m/s^2) + (60 kg * 9.8 m/s^2) = 1960 N.

Therefore, support 1 exerts an upward force of 1960 Newtons on the beam.

For support 2, we can apply the same reasoning. The sum of the vertical forces at support 2 must also be zero:

R2 - weight1 = 0.

Therefore, R2 is equal to the weight of the beam:

R2 = weight1 = 100 kg * 9.8 m/s^2 = 980 N.

Hence, support 2 exerts an upward force of 980 Newtons on the beam.