Simple Machines — Practice Quiz

7.5 kgf

Step 1: Identify arms — Load arm = 20 cm; Total length = 100 cm; so Effort arm = 100 − 20 = 80 cm. Step 2: Use the principle of lever — Load × Load arm = Effort × Effort arm. Step 3: Substitute — 60 × 20 = Effort × 80. Step 4: Solve — Effort = 1200 ÷ 80 = 15 kgf. Wait — re-check: the lever total length is 1 m = 100 cm, load is 20 cm from fulcrum (fulcrum at one end), so effort arm = 100 cm, load arm = 20 cm. Effort = (60 × 20) ÷ 100 = 1200 ÷ 100 = wait, but fulcrum is at end of shorter arm. So load arm = 20 cm, effort arm = 100 cm. Effort = (60 × 20) ÷ 100 = 12 kgf. Hmm — re-reading: 'fulcrum

Class II — load is between fulcrum and effort

Step 1: In a bottle opener, one end (tip) acts as the fulcrum, resting on the bottle cap edge. Step 2: The load (resistance of the cap) acts in the middle portion of the opener. Step 3: The effort is applied at the handle end, which is farthest from the fulcrum. Step 4: Since load is in the middle — between fulcrum and effort — it is a Class II lever. This is why bottle openers have a mechanical advantage greater than 1, making it easy to pop bottle caps.

Equal to 1, because load equals effort in an ideal pulley

Step 1: In an ideal (frictionless) single fixed pulley, the effort applied equals the load to be lifted. Step 2: MA = Load ÷ Effort = Load ÷ Load = 1. Step 3: So the pulley does NOT give a force advantage — its main benefit is changing the direction of effort (you pull downward to lift something upward). Step 4: In a real (actual) pulley, friction makes the effort slightly more than the load, so MA < 1. The usefulness of a pulley is direction change, not force multiplication.

40 kgf

Step 1: Recall — MA = Load ÷ Effort, so Effort = Load ÷ MA. Step 2: Substitute — Effort = 200 kgf ÷ 5. Step 3: Calculate — Effort = 40 kgf. Step 4: Verify — MA = 200 ÷ 40 = 5 ✓. Option A (1000 kgf) is wrong — that is Load × MA. Option B (195 kgf) is wrong — that is Load − MA. Option D (25 kgf) is a common error from wrong division.