Question

Diana raises a 1000. N piano a distance of 5.00 m using a set of pulleys. She pulls
in 20.0 m of rope. a) How much effort force did Diana apply if this was an ideal
machine? b) What force was used to overcome friction if the actual effort force
was 300. N? c) What was the work output? d) What was the ideal mechanical
advantage? e) What was the actual mechanical advantage, if the input force was
300N? +

Answer 1

Answer:

a) 250 N

b) 50 N

c) 5000 J

d) 4

e) 3.33

Explanation:

Given that weight of piano ($F_r$) = 1000 N, distance moved by pulley ($d_r$) = 5 m and the rope used ($d_e$) =20 m

a) The effort applied ($F_e$) if it was an ideal machine is:

$F_e=\frac{F_rd_r}{d_e} = \frac{1000*5}{20}=250N$

b) If the actual effort  = 300 N

The force to overcome friction = actual effort - $F_e$ = 300 - 250 = 50 N

c) Work output = $F_rd_r=1000*5=5000J$

d) ideal mechanical  advantage (IMA) = $\frac{d_e}{d_r} =\frac{20}{5} = 4$

e) Input force = 300 N, Therefore:

actual mechanical advantage = $F_r$ / input force = 1000 / 300 = 3.33

Answer 2

a. The amount of effort force Diana applied if this was an ideal  machine is 250 Newton.

b. The force which was used to overcome friction if the actual effort force was 300 is 50 Newton.

c. The work output of the set of pulleys is 5000 Nm.

d. The ideal mechanical  advantage (IMA) is 4.

e. The actual mechanical  advantage (AMA), if the input force was  300 Newton is 3.333.

Given the following data:

• Load force = 1000 N

• Load distance = 5.00 m

• Effort distance = 20.0 m

a. To determine the amount of effort force Diana applied if this was an ideal  machine:

For an ideal machine, the following equation relates load force with effort force;

$Effort\;force \times effort\;distance = Load\;force \times Load\;distance$

Substituting the given parameters into the formula, we have;

$Effort\;force = \frac{1000 \times 5.00}{20.0}$

Effort force = 250 Newton.

b. To determine the force which was used to overcome friction if the actual effort force was 300:

$Force = Actual\;effort - Effort\;force\\\\Force = 300-250$

Force = 50 Newton.

c. To calculate the work output:

$Work = Load \times load\;distance\\\\Work = 1000 \times 5.00$

Work output = 5000 Nm.

d. To calculate the ideal mechanical  advantage (IMA):

$IDA=\frac{Load\;force}{Effort\;force} \\\\IDA=\frac{1000}{250}$

Ideal mechanical  advantage (IMA) = 4.

e. To calculate the actual mechanical  advantage (AMA), if the input force was  300 Newton:

$AMA = \frac{Load\;force}{Inp. \;force} \\\\AMA = \frac{1000}{300}$

Actual mechanical  advantage (AMA) = 3.333

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