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This problem set deals with torque. |
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| 1. |
Circle the correct answer. Torque is defined as:
b. the product of the force applied and the length of the lever
arm.
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| 2. |
A force of 15 lb is used to produce a torque of 45.8 ft·lb. Find the
lever arm.
Note: I am substituting a capital L for lever arm because I don't feel
like trying to make a cursive L.
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| Knowns |
Unknowns |
Equations/Remarks |
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F = 15 lb
t = 45.8 ft·lb
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L = ? |
t = F·L |
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t = F·L
L = t/F
L = (45.8 ft·lb)/(15 lb)
= 3.05 ft
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Rearrange the original equation to solve for the unknown, which
in this case is the lever arm, L.
Substitute the given values and calculate.
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| 3. |
A force of 15 N is used to produce a torque of 130.5 N·m. Find the
lever arm.
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| Knowns |
Unknowns |
Equations/Remarks |
F = 15 N
t = 130.5 N·m |
L = ? |
t = F·L |
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t = F·L
L = t/F
L = (130.5 N·m)/(15 N)
= 8.7 m
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Rearrange the equation to solve for the unknown.
Substitute and calculate.
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| 4. |
A force of 9 lb is applied at right angles to a torque wrench. The
handle of the wrench is 18 in long.
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| A. |
What is the lever arm in feet? |
| B. |
Find the magnitude of the torque in
ft·lb. |
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| Knowns |
Unknowns |
Equations/Remarks |
F = 9 lb
L = 18 in
1 ft = 12 in |
A) L = ? in ft
B) t = ? in ft·lb |
t = F·L |
A) L = (18 in)(1
ft/12 in) = 1.5 ft
B) t = F·L
t = (9 lb)(1.5
ft) = 13.5 ft·lb |
Convert inches to feet
Substitute the known values and calculate the answer. |
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| 5. |
A uniform rod 6 ft long is balanced at its center. A weight of 5 lb
is suspended 1.25 ft from the center on the right. A second weight is
placed on the left 6 in from the center.
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| A. |
Draw a free body diagram of the rod showing the forces
as arrows. |
| B. |
Compute the clockwise torque. |
| C. |
What is the value of the counterclockwise torque?
Explain how you know. |
| D. |
How many pounds is the second weight? |
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| Knowns |
Unknowns |
Equation/Remarks |
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F1 = 5 lb
L1 = 1.25 ft right of center
L2 = 6 in left of center
1 ft = 12 in
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B) tCW
= ?
C) tCCW = ?
D) F2 = ? |
t = F·L |
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A) Free Body Diagram
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B) tCW
= F1·L1
= (5 lb)(1.25 ft) = 6.25 ft·lb CW
C) tCCW = 6.25
ft·lb CCW. The rod is balanced, therefore the sum of the torques
equals zero. This means that the two torques are equal in magnitude
but have opposite directions.
D) tCCW
= F2·L2
L2 = (6 in)(1
ft/12 in) = 0.5 ft
F2 = tCCW/L2
= (6.25 ft·lb)/(0.5 ft)
= 12.5 lb
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Substitute and calculate.
Rearrange the equation to solve for F2.
Note that 6 in needs to be converted to feet. Then substitute and
solve. |
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| 6. |
Four weights hang from a uniform 1 m long
rod which is supported at its center (50 cm). A weight of 30 N hangs at
10 cm, a 40 N weight hangs at 35 cm, and a 25 N weight hangs at 70 cm.
The rod is balanced (in equilibrium). The fourth weight is 50 N. |
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| A. |
Sketch and draw the free-body diagram. |
| B. |
Using torques and conditions for equilibrium,
find the location of the fourth weight. |
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This problem is a bit more complicated than the others, so I will explain
it first.
The pivot point of the rod is at a position of 50 cm as measured from
one end. I am assuming that the zero cm position is at the left end
and the 100 cm position is at the right end. I am using the following
sign conventions: all downward forces are positive, all positions to
the left of center have negative lever arms, all positions to the right
of center have positive lever arms, clockwise torques are positive,
and counterclockwise torques are negative.
The following variables are defined:
p = position in cm as measured from the left end of the rod
L = lever arm, measured from the pivot point
tNET = sum of all
the torques acting on the rod
The equation relating the position and the lever arm is p = L +
50 cm. The equation for the net torque is tNET = t1 + t2 + t3 + t4.
The rod is balanced, so the net torque acting on the rod must
be zero. Find the lever arms of the known positions using the equation
p = L + 50 cm. Then use the net torque equation
to find the lever arm of the fourth weight. Then use the position equation
to find the position of the fourth weight.
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| Knowns |
Unknowns |
Equations/Remarks |
F1 = 30 N
F2 = 40 N
F3 = 25 N
F4 = 50 N
p1 = 10 cm
p2 = 35 cm
p3 = 70 cm
tNET = 0
(equilibrium) |
p4 = ? |
p = L + 50 cm
tNET = t1 + t2 + t3 + t4
t = F·L |
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A) Free Body Diagram
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B) Solution
p = L + 50 cm
L = p - 50 cm
L1 = 10 cm - 50 cm = -40 cm
L2 = 35 cm - 50 cm = -15 cm
L3 = 70 cm - 50 cm = +20 cm
tNET = t1 + t2 + t3 + t4
t1 + t2 + t3 + t4
= 0
F1·L1
+ F2·L2
+ F3·L3
+ F4·L4
= 0
L4 = -(F1·L1
+ F2·L2
+ F3·L3)/(F4)
L4 = -[(30 N)(-40
cm) + (40 N)(-15 cm) + (25
N)(20 cm)]/(50 N)
L4 = 26 cm
p = L + 50 cm = 26 cm + 50 cm = 76 cm
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First, we must find the lever arms for the three known positions.
Then we use the torque equations to find the lever arm for the
fourth weight.
Finally, we use the position equation to find the position of
the fourth weight.
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| 7. |
When we use the English system of units, mechanical force is measured
in pounds. When we use the metric (or SI) units, mechanical force is
measured in
a. newtons
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In questions 8 - 12 use the following
vocabulary: vector, scalar, mass, weight, or torque. |
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| 8. |
A measure of the amount of matter contained
in an object is mass . |
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| 9. |
A physical quantity described by both magnitude
and direction is vector . |
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| 10. |
A physical quantity described only by magnitude
is scalar . |
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| 11. |
A measure of gravitational pull is weight . |
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