Question

Difficulty: MediumLevers and Fulcrums

A technician is testing a mechanical linkage that uses a straight lever pinned at one end, which serves as the fulcrum. A control rod is attached to the lever at a distance of 4 inches4\text{ inches} from the fulcrum and creates a resistance of 150 lbs150\text{ lbs}. An actuator applies an effort force at the free end of the lever, located 12 inches12\text{ inches} from the fulcrum. If all forces are applied perpendicular to the lever, what effort force (in pounds) is required to perfectly balance the control rod's resistance?

Answer: 50 lbs

Answer

50
In a balanced lever system, the torque produced by the effort force must equal the torque produced by the resistance force. Torque is calculated by multiplying the force by its distance from the fulcrum. Therefore, FE×12=150×4F_E \times 12 = 150 \times 4, which simplifies to 12FE=60012F_E = 600, resulting in an effort force of 50 lbs50\text{ lbs}.

Step-by-Step Solution

1
Identify the known variables and their distances from the fulcrum.
Resistance Force (FRF_R) = 150 lbs150\text{ lbs}, Resistance Arm (dRd_R) = 4 inches4\text{ inches}, Effort Arm (dEd_E) = 12 inches12\text{ inches}.
Correctly measuring distances from the pivot point (fulcrum) is necessary to calculate torque.
2
Set up the torque balance equation for the lever.
FE×dE=FR×dRF_E \times d_E = F_R \times d_R
For the lever system to be in equilibrium, the opposing torques must be equal.
3
Substitute the values and calculate the required effort force.
FE×12=150×4⇒12FE=600⇒FE=50 lbsF_E \times 12 = 150 \times 4 \Rightarrow 12F_E = 600 \Rightarrow F_E = 50\text{ lbs}.
Solving the algebraic equation provides the necessary balancing force.

Key Concept

Torque balance and lever mechanical advantage

Alternative Method

Calculate the mechanical advantage (MA) first. MA = Effort Arm ÷\div Resistance Arm = 12÷4=312 \div 4 = 3. Since the lever provides a mechanical advantage of 33, the required effort force is exactly one-third of the resistance force (150÷3=50 lbs150 \div 3 = 50\text{ lbs}).
Estimated Time:1m 0s
Rate this question