Question

a b c r_a r_b 0.2m 0.1m 10 nm

Explanation:

Step1: Identify the type of problem

This is a beam deflection or static equilibrium problem, likely involving moments and forces. We need to find the reactions \( R_A \) and \( R_B \) at supports A and B for the beam with a distributed load (or maybe a moment? Wait, the load is 10 Nm, which is a moment? Wait, no, 10 N/m would be a distributed load, but here it's 10 Nm, maybe a couple moment at C. Wait, the diagram: A is a pin support, B is a roller support. The distance from A to B is 0.2 m, B to C is 0.1 m. The load at C is 10 Nm (maybe a clockwise moment).

To find the reactions, we use equilibrium equations: sum of forces in y-direction, sum of moments about a point.

First, sum of forces in y-direction: \( R_A + R_B = 0 \)? Wait, no, maybe the load is a moment, not a force. Wait, if it's a moment (couple) at C, then the forces at A and B must form a couple to balance it. Wait, let's clarify the load: the diagram shows 10 Nm at C, probably a clockwise moment.

So, taking moments about A: the moment from \( R_B \) is \( R_B \times 0.2 \) (counterclockwise), and the moment at C is 10 Nm clockwise. The moment from \( R_A \) about A is zero (since it's at A). So equilibrium of moments: \( \sum M_A = 0 \)

\( R_B \times 0.2 - 10 = 0 \)? Wait, no, if the moment at C is clockwise, then the moment from \( R_B \) (upward force at B) about A is counterclockwise, so:

\( R_B \times 0.2 = 10 \) (since the clockwise moment at C must be balanced by the counterclockwise moment from \( R_B \))

Wait, no, let's do it properly. The sign convention: let's take counterclockwise as positive.

Moment about A: \( R_B \times 0.2 \) (counterclockwise) - 10 (clockwise, so negative) = 0

So \( 0.2 R_B - 10 = 0 \)

Then \( R_B = 10 / 0.2 = 50 \) N? Wait, but that seems odd. Wait, maybe the load is 10 N/m distributed load? Wait, the label is 10 Nm, which is a moment (unit of moment is Nm). So if it's a moment at C, then the force reactions: since there's no vertical force load, only a moment, the sum of vertical forces: \( R_A + R_B = 0 \), but that would mean \( R_A = -R_B \), and the moment from \( R_B \) about A is \( R_B \times 0.2 \), which must balance the moment at C (10 Nm clockwise). So \( R_B \times 0.2 = 10 \) (if moment at C is clockwise, then \( R_B \) must create a counterclockwise moment, so \( R_B \) is upward, \( R_A \) is downward).

Wait, maybe I misinterpret the load. Let's re-examine the diagram: the load is drawn as a curved arrow at C, labeled 10 Nm, so it's a clockwise moment (couple) at C.

So equilibrium equations:

1. Sum of forces in y-direction: \( R_A + R_B = 0 \) (since no vertical force loads, only the moment)
2. Sum of moments about A: \( R_B \times 0.2 - 10 = 0 \) (because \( R_B \) creates a counterclockwise moment about A, and the moment at C is clockwise, so they must balance)

From equation 2: \( R_B = 10 / 0.2 = 50 \) N (upward)

Then from equation 1: \( R_A = -R_B = -50 \) N (downward, meaning 50 N downward)

Wait, but that seems like a couple. Alternatively, maybe the load is 10 N/m distributed over BC? Wait, the label is 10 Nm, which is a moment, but maybe it's a typo and should be 10 N/m. Let's check both cases.

Case 1: Moment at C (10 Nm clockwise)

Forces: \( R_A \) (vertical at A), \( R_B \) (vertical at B)

Sum of forces: \( R_A + R_B = 0 \) (since no vertical forces, only moment)

Sum of moments about A: \( R_B \times 0.2 - 10 = 0 \) → \( R_B = 50 \) N, \( R_A = -50 \) N

Case 2: Distributed load 10 N/m over BC (length 0.1 m), so total force is \( 10 \times 0.1 = 1 \) N, acting at the midpoint of BC, which is 0.2 + 0.05 = 0.25 m from A.

Then sum of forces: \( R_A + R_B = 1 \) N

Sum of moments about A: \( R_B \times 0.2 - 1 \times 0.25 = 0 \) → \( 0.2 R_B = 0.25 \) → \( R_B = 1.25 \) N, \( R_A = 1 - 1.25 = -0.25 \) N. But the label is 10 Nm, so probably a moment.

Assuming it's a moment (10 Nm clockwise at C):

Step1: Sum of moments about A

\( \sum M_A = 0 \)

Moment from \( R_B \): \( R_B \times 0.2 \) (counterclockwise, positive)

Moment at C: -10 (clockwise, negative)

So \( 0.2 R_B - 10 = 0 \)

Step2: Solve for \( R_B \)

\( 0.2 R_B = 10 \)

\( R_B = \frac{10}{0.2} = 50 \) N (upward)

Step3: Sum of forces in y-direction

\( \sum F_y = 0 \)

\( R_A + R_B = 0 \) (since no vertical force loads)

\( R_A = -R_B = -50 \) N (downward, so 50 N downward)

Wait, but maybe the support at A is a pin (can have vertical and horizontal reactions), but the load is vertical? No, the load is a moment, so horizontal reaction at A would be zero. So vertical reactions: \( R_A = -50 \) N (downward), \( R_B = 50 \) N (upward)

Alternatively, if the load is 10 N/m (distributed load) over BC (length 0.1 m), then total load is \( 10 \times 0.1 = 1 \) N, acting at 0.2 + 0.05 = 0.25 m from A.

Then sum of moments about A: \( R_B \times 0.2 - 1 \times 0.25 = 0 \) → \( R_B = (1 \times 0.25)/0.2 = 1.25 \) N

Sum of forces: \( R_A + R_B = 1 \) → \( R_A = 1 - 1.25 = -0.25 \) N

But the label is 10 Nm, so more likely a moment.

So the reactions are \( R_A = 50 \) N downward, \( R_B = 50 \) N upward (forming a couple to balance the 10 Nm moment? Wait, no, the distance between A and B is 0.2 m, so the moment from \( R_A \) and \( R_B \) is \( R_B \times 0.2 - R_A \times 0 = R_B \times 0.2 \) (since \( R_A \) is at A). Wait, if \( R_A \) is downward and \( R_B \) is upward, the moment between them is \( (R_B - R_A) \times 0.2 \)? No, moment about A is \( R_B \times 0.2 \) (since \( R_A \) is at A, its moment is zero). So to balance a 10 Nm clockwise moment, we need a 10 Nm counterclockwise moment from \( R_B \), so \( R_B \times 0.2 = 10 \) → \( R_B = 50 \) N, and \( R_A = -50 \) N (downward) to satisfy force equilibrium (since there's no vertical force, sum of vertical forces must be zero).

Answer:

If we need to find \( R_B \), the answer is \( \boxed{50} \) N (upward). If \( R_A \), it's \( \boxed{-50} \) N (or 50 N downward). But likely the question is to find \( R_B \), so 50 N.