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Machines - Simple Machines

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A machine is a device that helps us do work by changing the magnitude, direction, or point of application of a force.

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Work Done (WW): In physics, work is done when a force FF causes a displacement ss in the direction of the force. It is given by W=F×sW = F \times s. The SI unit is the Joule (JJ).

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Mechanical Advantage (MAMA): It is the ratio of the load LL to the effort EE. If MA>1MA > 1, the machine acts as a force multiplier. It is expressed as MA=LEMA = \frac{L}{E}.

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Velocity Ratio (VRVR): It is the ratio of the distance moved by the effort dEd_E to the distance moved by the load dLd_L. It is given by VR=dEdLVR = \frac{d_E}{d_L}.

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Efficiency (η\eta): The ratio of useful work output to the total work input. For an ideal machine, η=100%\eta = 100\%, but in reality, η<100%\eta < 100\% due to friction and the weight of moving parts.

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Relationship: The three variables are linked by the formula MA=VR×ηMA = VR \times \eta.

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Levers: Classified into three types based on the position of the Fulcrum (FF), Load (LL), and Effort (EE). In Class I, the fulcrum is in the middle; in Class II, the load is in the middle; in Class III, the effort is in the middle.

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Pulleys: A single fixed pulley has a VR=1VR = 1 and only changes the direction of force. A single movable pulley has a VR=2VR = 2. For a block and tackle system with nn pulleys, VR=nVR = n.

📐Formulae

Work(W)=F×sWork (W) = F \times s

Mechanical Advantage (MA)=Load (L)Effort (E)Mechanical\ Advantage\ (MA) = \frac{Load\ (L)}{Effort\ (E)}

Velocity Ratio (VR)=Distance moved by Effort (dE)Distance moved by Load (dL)Velocity\ Ratio\ (VR) = \frac{Distance\ moved\ by\ Effort\ (d_E)}{Distance\ moved\ by\ Load\ (d_L)}

Efficiency (η)=Work OutputWork Input×100%Efficiency\ (\eta) = \frac{Work\ Output}{Work\ Input} \times 100\%

Efficiency (η)=MAVREfficiency\ (\eta) = \frac{MA}{VR}

Principle of Moments: Effort×Effort Arm=Load×Load ArmPrinciple\ of\ Moments:\ Effort \times Effort\ Arm = Load \times Load\ Arm

💡Examples

Problem 1:

A crowbar of length 1.5 m1.5\ m is used as a Class I lever to lift a stone of weight 600 N600\ N. The fulcrum is placed 0.3 m0.3\ m from the stone. Calculate the effort required to lift the stone, assuming 100%100\% efficiency.

Solution:

Given: Load L=600 NL = 600\ N. Load Arm dL=0.3 md_L = 0.3\ m. Total length = 1.5 m1.5\ m. Therefore, Effort Arm dE=1.5 m−0.3 m=1.2 md_E = 1.5\ m - 0.3\ m = 1.2\ m. Using the principle of moments: E×dE=L×dLE \times d_E = L \times d_L E×1.2=600×0.3E \times 1.2 = 600 \times 0.3 E=1801.2=150 NE = \frac{180}{1.2} = 150\ N

Explanation:

The effort required is 150 N150\ N. Since the effort arm is longer than the load arm, the machine acts as a force multiplier.

Problem 2:

A pulley system has a velocity ratio of 44. It is used to lift a load of 1000 N1000\ N by applying an effort of 312.5 N312.5\ N. Calculate the efficiency of the machine.

Solution:

Given: VR=4VR = 4, L=1000 NL = 1000\ N, E=312.5 NE = 312.5\ N. First, find Mechanical Advantage (MAMA): MA=LE=1000312.5=3.2MA = \frac{L}{E} = \frac{1000}{312.5} = 3.2 Now, find Efficiency (η\eta): η=MAVR×100%\eta = \frac{MA}{VR} \times 100\% η=3.24×100%=0.8×100%=80%\eta = \frac{3.2}{4} \times 100\% = 0.8 \times 100\% = 80\%

Explanation:

The efficiency is 80%80\%. The 20%20\% loss is likely due to friction in the pulleys and the weight of the strings/pulleys.