Question

part 9 - mechanical comprehension question 13 of 1
if the beam has a heavy load placed upon it at the point shown, about how much downward force is exerted on support a?
a 25 lb.
b 50 lb.
c 75 lb.
d 100 lb.
click the button or type the letter next to your answer.

Explanation:

Step1: Analyze the beam's segments

The beam is divided into 4 equal segments (from A to the load is 1 segment, load to B is 3 segments? Wait, no, looking at the diagram: the load is at the 2nd mark from A (A to first mark: 1, first to load: 1, so load is 1/4 from A? Wait, no, the beam has marks: A, then 4 marks? Wait, the supports are A and B, with the beam having 4 intervals (since 5 marks: A, then 4 spaces to B). The load is at the 2nd mark (1st interval from A). So the distance from A to load is 1 unit, from load to B is 3 units? Wait, no, let's use the principle of moments (torque equilibrium). For equilibrium, the moment about B should be zero. Moment = force × distance. Let the force at A be \( F_A \), force at B be \( F_B \). The load is 100 lb, distance from load to B: let's say each segment is length \( L \). So distance from A to load is \( L \), from load to B is \( 3L \), and from A to B is \( 4L \).

Step2: Apply torque equilibrium about B

Torque about B: \( F_A \times 4L - 100 \times 3L = 0 \)? Wait, no, wait: the load is on the left side of the beam, closer to A. Wait, maybe I got the distances wrong. Let's count the segments: from A to the load is 1 segment, load to B is 3 segments? Wait, the diagram shows the load is at the 2nd mark (A is at 0, first mark at 1, load at 2, then 3, 4, B at 5? No, the beam has A, then four intervals (so five points: A, 1, 2, 3, 4, B? Wait, the image shows the beam with A, then a mark, then the load, then three more marks to B. So from A to load: 1 interval, load to B: 3 intervals. So length from A to load: \( d_1 = 1 \), load to B: \( d_2 = 3 \), total length A to B: \( d = 4 \).

Torque about B: \( F_A \times d - 100 \times d_2 = 0 \) (since torque from \( F_A \) is clockwise, load is counterclockwise? Wait, no: if we take moments about B, the force at A creates a clockwise moment (since it's to the left of B), and the load creates a counterclockwise moment (to the left of B, so force down, moment about B is load × distance from load to B (counterclockwise), and \( F_A \) is up, moment about B is \( F_A \times \) distance from A to B (clockwise). Wait, no, forces: the load is downward (100 lb), \( F_A \) and \( F_B \) are upward. So for equilibrium, sum of forces: \( F_A + F_B = 100 \). Sum of moments about A: \( 100 \times 1L - F_B \times 4L = 0 \) (since load is 1L from A, B is 4L from A). So \( 100L = F_B \times 4L \) → \( F_B = 100 / 4 = 25 \) lb? Wait, no, that can't be. Wait, maybe the distance from A to load is 3L, and load to B is 1L? Wait, maybe I flipped the distances. Let's look again: the load is closer to A? Wait, the diagram: A is on the left, load is on the left part of the beam, so distance from A to load is 1 segment, load to B is 3 segments. Wait, no, maybe the load is at 1/4 from A, so distance from A to load is 1, load to B is 3. Then moment about A: \( 100 \times 1 - F_B \times 4 = 0 \) → \( F_B = 25 \), then \( F_A = 100 - 25 = 75 \)? Wait, that contradicts. Wait, no, maybe the load is at 3/4 from A, so distance from A to load is 3, load to B is 1. Then moment about A: \( 100 \times 3 - F_B \times 4 = 0 \) → \( F_B = 75 \), then \( F_A = 25 \). Ah, that makes sense! Because the load is closer to B? Wait, no, the diagram shows the load is on the left side, near A. Wait, maybe I misread the diagram. Let's see: the load is at the second mark from A (A, then a mark, then the load, then three marks to B). So from A to load: 1 interval, load to B: 3 intervals. So distance from A to load: \( x \), load to B: \( 3x \), total length \( 4x \).

Moment about A: the load creates a clockwise moment (force down, distance \( x \) from A), and \( F_B \) creates a counterclockwise moment (force up, distance \( 4x \) from A). So equilibrium: \( 100 \times x = F_B \times 4x \) → \( F_B = 100 / 4 = 25 \)? No, that would mean \( F_A = 100 - 25 = 75 \). Wait, that's the opposite. Wait, maybe the load is at 3 intervals from A, 1 from B. Let's assume the beam is divided into 4 equal parts, with the load at the 3rd part from A (i.e., 1 part from B). So distance from A to load: \( 3L \), load to B: \( L \). Then moment about A: \( 100 \times 3L = F_B \times 4L \) → \( F_B = (100 \times 3L) / 4L = 75 \) lb. Then \( F_A = 100 - 75 = 25 \) lb? No, that's not right. Wait, no, the force at A and B are upward, load is downward. So sum of forces: \( F_A + F_B = 100 \). Sum of moments about B: \( F_A \times 4L = 100 \times 3L \) (if load is 3L from B). Then \( F_A = (100 \times 3L) / 4L = 75 \) lb. Ah, that makes sense. Wait, maybe I had the distance from load to B wrong. Let's look at the diagram again: the load is on the left side, so from A to load is 1 segment, load to B is 3 segments. So distance from A to B is 4 segments. So moment about B: the force at A creates a clockwise moment (since it's to the left of B, force up, moment is \( F_A \times 4 \) (length A to B)), and the load creates a counterclockwise moment (force down, distance from load to B is 3 segments, so moment is \( 100 \times 3 \)). For equilibrium, clockwise moment = counterclockwise moment: \( F_A \times 4 = 100 \times 3 \) → \( F_A = (100 \times 3)/4 = 75 \) lb. Wait, that's 75 lb. So the force at A is 75 lb.

Answer:

C. 75 lb.