Machines — ICSE Class 10 Physics Important Questions

Answer: (b) changes the direction of the effort.

A single fixed pulley has $VR=1$VR=1 and ideal $MA=1$MA=1, so it does not reduce the effort; its use is to change the direction of the applied effort, making work more convenient.

Answer: (b) class II.

When the load is between the fulcrum (wheel axle) and the effort (hands), it is a class II lever, which always has $MA>1$MA>1.

Answer: (b) $4$4.

$\eta=\dfrac{MA}{VR}\Rightarrow MA=\eta\times VR=0.80\times5=4.$η=MA/VR⇒ MA=η × VR=0.80×5=4.

Answer: (b) $100\text{ N}$100 N.

The two strands together support the load and the pulley: $2E=180+20=200\text{ N}$2E=180+20=200 N, so $E=100\text{ N}$E=100 N. The mechanical advantage is then $\dfrac{180}{100}=1.8$180/100=1.8, which is less than $VR=2$VR=2 even without friction.

Answer: (a) Since $\eta=\dfrac{MA}{VR}$η=MA/VR and friction makes $\eta<1$η<1, we get $MA<VR$MA<VR for a real machine. R correctly explains A.

Mechanical advantage (MA) is the ratio of the load to the effort: $MA=\dfrac{L}{E}$MA=L/E.

Velocity ratio (VR) is the ratio of the distance moved by the effort to the distance moved by the load in the same time: $VR=\dfrac{d_E}{d_L}$VR=d_E/d_L.

Both are ratios of two like quantities (force/force and distance/distance), so the units cancel and they are pure numbers with no units.

Effort arm $=1.5-0.25=1.25\text{ m}$=1.5-0.25=1.25 m; load arm $=0.25\text{ m}$=0.25 m.

For an ideal lever, $MA=VR=\dfrac{\text{effort arm}}{\text{load arm}}$MA=VR=(effort arm)/(load arm):
$MA=\frac{1.25}{0.25}=5.$MA=1.25/0.25=5.

The mechanical advantage is $5$5.

In a single movable pulley the load hangs from the movable pulley and two strands of the string support it; the effort is applied to the free end.

$VR=2$VR=2 (two strands support the load, and $VR$VR does not depend on friction). Ideal $MA=2$MA=2; in practice $MA$MA is less than $2$2 because of friction and the weight of the pulley.

Advantage: the effort needed is about half the load. Disadvantage: the effort has to be applied in the upward direction, which is inconvenient.

(i) Mechanical advantage:
$MA=\frac{L}{E}=\frac{600}{200}=3.$MA=L/E=600/200=3.

(ii) Velocity ratio:
$VR=\frac{d_E}{d_L}=\frac{8}{2}=4.$VR=d_E/d_L=8/2=4.

(iii) Efficiency:
$\eta=\frac{MA}{VR}=\frac{3}{4}=0.75=75\%.$η=MA/VR=3/4=0.75=75%.

The machine has $MA=3$MA=3, $VR=4$VR=4 and efficiency $75\%$75%.

Velocity ratio $VR=5$VR=5, efficiency $\eta=0.60$η=0.60.

Mechanical advantage:
$MA=\eta\times VR=0.60\times5=3.$MA=η × VR=0.60×5=3.

Effort:
$E=\frac{L}{MA}=\frac{150}{3}=50\text{ N}.$E=L/MA=150/3=50 N.

Effort distance $=VR\times$=VR× load distance $=5\times4=20\text{ m}$=5×4=20 m.

Work input $=E\times d_E=50\times20=1000\text{ J}$=E× d_E=50×20=1000 J.
Useful work output $=L\times d_L=150\times4=600\text{ J}$=L× d_L=150×4=600 J.

Work done against friction $=$= input $-$- output $=1000-600=400\text{ J}$=1000-600=400 J.

Effort $=50\text{ N}$=50 N and work lost to friction $=400\text{ J}$=400 J.

Let an effort $E$E move through a distance $d_E$d_E while the machine raises a load $L$L through a distance $d_L$d_L.

Work input $=E\times d_E$=E× d_E and useful work output $=L\times d_L$=L× d_L.

$\eta=\frac{\text{output}}{\text{input}}=\frac{L\times d_L}{E\times d_E}=\frac{L}{E}\times\frac{d_L}{d_E}=MA\times\frac{1}{VR}.$η=output/input=(L× d_L)/(E× d_E)=L/E× d_L/d_E=MA× 1/VR.

So $\eta=\dfrac{MA}{VR}$η=MA/VR, or $MA=\eta\times VR$MA=η × VR. For an ideal machine $\eta=1$η=1, so $MA=VR$MA=VR.

(a) A lever is a rigid bar capable of turning about a fixed point called the fulcrum, on which a load is overcome by an effort. The three orders are shown below.

Class I (fulcrum in the middle): a see-saw. Class II (load in the middle): a wheelbarrow. Class III (effort in the middle): a pair of forceps.

(b) Scissors are class I levers, with the rivet as the fulcrum.

• Cloth scissors: the load arm (blade) is longer than the effort arm (handle), so $MA<1$MA<1. Cloth offers little resistance, so a large force is not needed; instead the long blades move much further than the fingers, giving a gain in speed and a longer cut with each stroke.
• Metal shears: the effort arm (handle) is longer than the load arm (blade), so $MA>1$MA>1. A small effort produces the large force needed to cut a metal sheet.

In this block and tackle, $5$5 strands of the string support the load, so $VR=5$VR=5.

(i) Velocity ratio $VR=$VR= number of load-supporting strands $=5$=5.

(ii) Mechanical advantage:
$MA=\frac{L}{E}=\frac{200}{50}=4.$MA=L/E=200/50=4.

(iii) Efficiency:
$\eta=\frac{MA}{VR}=\frac{4}{5}=0.8=80\%.$η=MA/VR=4/5=0.8=80%.

(iv) For an ideal machine $MA=VR=5$MA=VR=5, so
$E_{\text{ideal}}=\frac{L}{VR}=\frac{200}{5}=40\text{ N}.$E_ideal=L/VR=200/5=40 N.

An ideal frictionless system would need only $40\text{ N}$40 N.

(i) The effort (biceps) lies between the fulcrum (elbow) and the load (hand), so the forearm is a class III lever.

(ii) For an ideal lever $VR=\dfrac{\text{effort arm}}{\text{load arm}}=\dfrac{4}{32}=\dfrac{1}{8}=0.125.$VR=(effort arm)/(load arm)=4/32=1/8=0.125.

(iii) Since the ideal $MA=VR=0.125<1$MA=VR=0.125<1, the mechanical advantage is less than $1$1. This means the muscle must exert a force greater than the load (about $8$8 times the load) to hold it.

(iv) Although it needs a large effort, the arrangement gives a large gain in speed and range of movement: a small movement of the biceps produces a large, fast movement of the hand.