A uniform beam of length l whose mass is 3.5kg rests with its ends on two digital scales. A block with a mass of 1.3kg rests on the beam, 1/4 of the distance in from the left balance. What do the scales read?

To find out what the scales read, let's break down the problem step by step.

Step 1: Calculate the center of mass of the beam.
The beam is uniform, so the center of mass lies at the midpoint. Hence, the center of mass of the beam is l/2.

Step 2: Calculate the position of the block.
The block is placed 1/4 of the distance from the left end of the beam. So, the position of the block is (1/4) * l.

Step 3: Calculate the distances from the left end to the block and from the right end to the block.
The distance from the left end to the block is given as (1/4) * l.
The distance from the right end to the block is given as (3/4) * l.

Step 4: Calculate the weight of the beam.
The weight of the beam is given by the formula W_beam = m_beam * g,
where m_beam is the mass of the beam and g is the acceleration due to gravity (9.8 m/s^2).
W_beam = 3.5 kg * 9.8 m/s^2 = 34.3 N.

Step 5: Calculate the weight of the block.
The weight of the block is given by the formula W_block = m_block * g,
where m_block is the mass of the block and g is the acceleration due to gravity (9.8 m/s^2).
W_block = 1.3 kg * 9.8 m/s^2 = 12.74 N.

Step 6: Calculate the weight supported by each scale.
Since the beam is in equilibrium, the sum of the clockwise torques about any point must be equal to the sum of the counterclockwise torques.
The weight supported by the left scale is given by W_left = (W_beam * distance_to_block_from_left) / length_of_beam,
where distance_to_block_from_left is (1/4) * l, and length_of_beam is l.
W_left = (34.3 N * (1/4) * l) / l = 8.575 N.

The weight supported by the right scale is given by W_right = W_beam - W_left.
W_right = 34.3 N - 8.575 N = 25.725 N.

Step 7: Determine the readings on each scale.
The left scale reads 8.575 N, and the right scale reads 25.725 N.

To find out what the scales read, we can start by considering the overall forces acting on the beam. Since the beam is in equilibrium (not moving), the sum of all forces acting on it must be equal to zero.

Let's call the left scale reading F1 and the right scale reading F2.

1. First, consider the forces acting vertically:

- The weight of the beam itself. Since it is a uniform beam, we can assume its weight is evenly distributed along its length. So, the weight acts at the center of the beam, which is at half of its length. The weight force is given by: F_beam = (mass of beam) x (acceleration due to gravity).

F_beam = 3.5 kg x 9.8 m/s^2. (Note: The beam's weight will be evenly distributed on both scales, so we will divide this force equally between F1 and F2 later)

- The weight of the block. It acts vertically downward at a distance 1/4 of the total length of the beam from the left balance. Let's call this distance x. The weight force of the block is given by: F_block = (mass of block) x (acceleration due to gravity).

F_block = 1.3 kg x 9.8 m/s^2.

- Forces acting on the scales. These are the forces being measured by the scales.

Now, let's set up the equilibrium condition considering the vertical forces:

F1 + F2 + F_beam + F_block = 0.

2. Consider the horizontal forces:

Since the beam is in equilibrium, the sum of the torques (moments) acting on it must also be zero.

- The beam's weight creates a clockwise torque around the left balance and a counterclockwise torque around the right balance. These torques are equal in magnitude.

To find the torques, we multiply the force by the distance from the pivot point (balance).

The clockwise torque created by the beam's weight is:
Torque_clockwise = (F_beam / 2) x (l/2),

while the counterclockwise torque created by the beam's weight is:
Torque_counterclockwise = (F_beam / 2) x (l/2).

- The block's weight creates a clockwise torque around the left balance.

The torque created by the block's weight is:
Torque_block = F_block x x.

Now, let's set up the equilibrium condition considering the horizontal torques:

Torque_clockwise + Torque_block = Torque_counterclockwise.

3. Solving the equations:

From the equilibrium condition for vertical forces, we know that F1 + F2 + F_beam + F_block = 0.

Since the beam's weight is evenly distributed on both scales, we can divide it between F1 and F2:
F_beam = 3.5 kg x 9.8 m/s^2 / 2 = 17.15 N

Therefore, our equation becomes:
F1 + F2 + 17.15 N + (1.3 kg x 9.8 m/s^2) = 0.

From the equilibrium condition for horizontal torques, we know that Torque_clockwise + Torque_block = Torque_counterclockwise.

Substituting the torques:
(F_beam / 2) x (l/2) + (F_block x x) = (F_beam / 2) x (l/2).

Simplifying, we get:
(F_block x x) = 0.

Since there is no force acting horizontally on the right scale, F2 = 0.

Finally, solving the equations will give us the scale readings:
F1 = - (17.15 N + (1.3 kg x 9.8 m/s^2)).
F2 = 0.

Note: The negative sign on F1 indicates it is pointing down while F2 points up.