Question

10n a c b d 10n·m 1m 1m 1m

Explanation:

Step1: Identify Supports and Loads

The beam has a pin support at \( A \) (which can provide horizontal and vertical reactions, and a moment reaction? Wait, no—pin support provides horizontal and vertical forces, but no moment. Wait, the diagram shows a pin at \( A \) (two reaction components) and a roller at \( B \) (vertical reaction only). The loads are a vertical force of \( 10 \, \text{N} \) at \( C \) (1 m from \( A \)), and a couple moment of \( 10 \, \text{N·m} \) at \( D \) (1 m from \( B \), so total length from \( A \) to \( D \) is \( 1 + 1 + 1 = 3 \, \text{m} \)).

Step2: Calculate Reactions at \( A \) and \( B \)

First, sum the vertical forces. Let \( R_{Ay} \) be the vertical reaction at \( A \), \( R_{By} \) at \( B \).

$$ \sum F_y = 0 \implies R_{Ay} + R_{By} - 10 = 0 \quad (1) $$

Sum the moments about \( A \). The moment due to \( 10 \, \text{N} \) at \( C \) is \( 10 \times 1 = 10 \, \text{N·m} \) (clockwise), the moment due to \( R_{By} \) is \( R_{By} \times 2 \) (counterclockwise, since \( B \) is 2 m from \( A \)), and the couple moment at \( D \) is \( 10 \, \text{N·m} \) (counterclockwise? Wait, the diagram shows the couple as a right arrow and left arrow, which is a counterclockwise moment? Wait, no—standard couple moment: if the arrows are as drawn (right at top, left at bottom at \( D \)), that's a counterclockwise moment. Wait, actually, the direction: a couple with a top arrow to the right and bottom arrow to the left creates a counterclockwise moment. So moment about \( A \):

$$ \sum M_A = 0 \implies -10 \times 1 + R_{By} \times 2 + 10 = 0 $$

Simplify: \( -10 + 2R_{By} + 10 = 0 \implies 2R_{By} = 0 \implies R_{By} = 0 \)
Then from equation (1): \( R_{Ay} + 0 - 10 = 0 \implies R_{Ay} = 10 \, \text{N} \) (upward)

Now, check horizontal reaction at \( A \): since there are no horizontal loads, \( R_{Ax} = 0 \).

Step3: Verify with Moment about \( B \)

Moment about \( B \): \( R_{Ay} \times 2 - 10 \times 1 - 10 = 0 \) (since \( R_{Ay} \) is upward, moment about \( B \) is \( 10 \times 2 = 20 \), \( 10 \, \text{N} \) at \( C \) is \( 10 \times 1 = 10 \) (clockwise), couple moment at \( D \) is \( 10 \) (clockwise? Wait, no—wait, maybe I messed up the couple direction. Wait, maybe the couple is clockwise? Let's re-examine the diagram. The couple at \( D \): top arrow right, bottom arrow left—actually, that's a counterclockwise moment. Wait, no—if you have a beam, and at \( D \), top fiber is pulled right, bottom fiber pulled left, that causes a counterclockwise rotation. But when calculating moment about \( B \), the distance from \( B \) to \( D \) is 1 m. So moment about \( B \):

$$ \sum M_B = 0 \implies R_{Ay} \times 2 - 10 \times (2 - 1) - 10 = 0 \implies 2R_{Ay} - 10 - 10 = 0 \implies 2R_{Ay} = 20 \implies R_{Ay} = 10 \, \text{N} $$

Which matches. So reactions: \( R_{Ax} = 0 \), \( R_{Ay} = 10 \, \text{N} \) (up), \( R_{By} = 0 \).

Answer:

Reactions: \( R_{Ax} = 0 \, \text{N} \), \( R_{Ay} = 10 \, \text{N} \) (upward), \( R_{By} = 0 \, \text{N} \)