QUIZ 13 - CH 11 - STATICS---ROTATIONAL AND TRANSLATIONAL EQUILIBRIUM, STRESS , STRAIN AND YOUNG'S MODULUS, SHEER FORCES |
3**, 11**, 10** (ladder--see example 11.3**), 12**, 13**,
14**, 18**, 24** (tensile tress,
strain---try 25**), 35** ( bulk modulus), 36** ( shear stress) 51**, 50** . |
DISCUSSIONS: ON THE FINAL EXAM YOU WILL LIKELY GET A PROBLEM INVOLVING A BEAM AND CABLE. Note: I mentioned that I may ask you to compute the instantaneous torque and angular acceleration if the cable was severed ( or not there in he first place) and the beam was free to swing about a given axis. YOU COULD ALSO GET A HORIZONTAL BEAM (PLANK) PROBLEM AS DISPLAYED BY # 3, 11 and12. SO PAY ATTENTION. There could be qualitative scantron questions on all topics. |
In the beam and ladder problems I tend to choose the axis at the left end of beam /ladder be it raised or lowered relative to the other end. That way we gain a consistency and lose none of the general principles in situations requiring discipline of the mind. |
3.***The net torque on the plank (beam) is zero about any axis and
in particular the axis shown at the centered pivot point. You can
think of all the mass of the plank being concentrated at the center
of mass (cm) of plank. The san exerts a force that applied at the
center of the sand, and thus is 0.375 m from the left
end. We have sum of torques about central axis D*mg - D'*Mg = 0, where m is unknown, D = 1 m - 0.375 m = 0.625 m, D' = 0.50 m, and M = 25.0 kg . Find m. NOTE YOU CAN ALSO FIND THE NORMAL FORCE OF MAGNITUDE N ACTING UPWARD AT THE PIVOT POINT: sum of the forces in y direction = 0 = pos - neg = N = mg - Mg. Solve for N! |
11.*** When you sum the torques to zero choose the axis at the
support point 2.00 m from the diver who is about to do a back
flip. Note that force on beam at the left end points down and had
magnitude N. For torques we have about the axis at given support
point: (1.00 m)*N - (0.50 m)* Mg - (2.00 m)*mg = 0. Find N. Note we must include the weight of the beam of mass M which acts downward at the beam midpoint a distance of 0.50 m from the support point. This allows you to answer (b) (a) sum of forces in the y direction = 0 = pos - beg = N' - N - mg - Mg. Find N' = magnitude of upward normal force at support point,. |
10***. In the beam and ladder problems I tend to choose
the axis at the left end of beam /ladder be it raised or
lowered relative to the other end. That way we gain a consistency
and lose none of the general principles in situations
requiring discipline of the mind. See example 11. 3. Because
of my choice of axis, I solve for N, the normal force of the
vertical wall on the ladder's top, a force that points
leftward. The ladder and wall form a 3-4-5 right triangle.
The top of the ladder ( right end) is 4 meters above the ground. I
choose the axis at the bottom of the ladder, i.e. at the ladder's
left end, which is 3.0 m from the wall. Sum of torques about axis
at ground = 0 = (4 m)*N - (L/2)*cos 53*Mg - (L/5)*cos 53*mg = 0,
where clearly M is the beam mass and m is the person mass (They
said he was L/5 the way up beam of length L.) Note M
<m numerically. From this you can find the normal force of magnitude N exerted by the wall on the ladder's top end . (That force points left as in figure 11.9) This allows you to solve for the friction force 0 = f_{s} - N in part (b). To get the maximum friction force, find N' the magnitude of the normal force on ladder from ground by simply summing the vertical forces to zero = N' - mg -Mg, or N' = (m +M)*g. Use the formula for the maximum static friction force and you have part (a). As the person climbs upward , we can see the dependence f_{s} has on D, the distance the person climbs up from the bottom. If f_{s} gets bigger than the value you found in part (a), then you have a problem. Using the torque equation and swapping out N for f_{s} we get 4*f_{s} = (L/2)*cos 53*Mg + D*cos 53*mg . Find for part (c) D if f_{s} is the maximum allowed before slipping. |
3**, 11**, 10** (ladder--see example 11.3**), 12**, 13**,
14**, 18**, 24** (tensile tress,
strain---try 25**), 35** ( bulk modulus), 36** ( shear stress) 51**, 50** . |
12.*** (b) This is perhaps the most intriguing problem of the bunch. Let's answer this conceptually. The man cannot walk too far to the right. If he exceeds a certain distance, the center of mass of the system defined by him and the beam will be to the right of the pivot point B, which means the beam would rotate clockwise. Think of it this way, when the system center of mass is between the pivot points you can choose one pivot points as an axis of rotation , say B, and you will find that that the counter clockwise torque by the concentrated mass m + M located at the center of mass will be counteracted by a clockwise torque from the upward normal force exerted by the other pivot A. But if that system center of mass just goes to the right of pivot B the beam will rotate clockwise---due to the torque from concentrated mass m + M---- unopposed by a countervailing torque and since the beam's left end lifts up from pivot point A instead of being forced downward into it . In that condition the beam begins to rotate clockwise because the center of mass is located at pivot B. You can find x to the right of B such that the cm is at B. EASIER WAY: There is an easier way to do it. Just look at the beam by its self as I did in class, and find the forces on beam from the boy, its own weight and the normal forces at the pivot points. Choose the axis of rotation to be B. That eliminates the normal force at B and includes only N, the magnitude of the normal force at A. Let the mass be a distance x to the right of B. The net torque on the beam is zero: 0 = (2. 5 m)*Mg - (5.00 m)*N - x*mg. Note we have placed the center of mass of the beam at the center of beam, 2.5 m from pivot B. When the beam starts to rotate, the normal N = 0. Find x. Note in this case, the cm is located at pivot B. (c) THIS PART IS PURE GEOMETRY. When the boy is at the end, the distance from the pivot point is x. At the same time, the distance of the center of mass of the beam from the pivot point is D. If the system center of mass is located at pivot B, we have the same condition as in part (b): The normal force at the other pivot point (A) is zero. Thus 0 = D*Mg - x*mg. At the same time you have another equation: D + x = 4.5, since the beam cm is located at the geometric center of the 9.00 m long beam. Solve these two equations for D and x. |
3**, 11**, 10** (ladder--see example 11.3**), 12**, 13**,
14**, 18**, 24** (tensile tress,
strain---try 25**), 35** ( bulk modulus), 36** ( shear stress) 51**, 50** . |
13. (a) Reviewed as an ICQ. Axis at lower left end of beam: 0
= L*sin30*T - (L/2)*cos 30*Mg -(L)*cos30*mg gives you T;
note m = M according to the problem. Solve for F_{h}, the horizontal component of force on beam from pivot. F_{h} = T. Solve for F_{v}, the vertical component of force on beam from pivot: 0 = F_{v} - mg - Mg, so F_{v} = (m + M)*g . (b) Axis at lower left end of beam: 0 = L*sin15*T - (L/2)*cos 45*Mg -(L)*cos45*mg gives you T; note m = M according to the problem. Solve for F_{h}, the horizontal component of force on beam from pivot. F_{h} = T*cos30 Solve for F_{v}, the vertical component of force on beam from pivot: 0 = F_{v} - mg - Mg - T*sin 30, so F_{v} = (m + M)*g + T*sin 30. |
14. This problem repeats what I did in class. Choose axis at the
beam left end. Find T using the torque equation. Draw a perpendicular from the axis to the line of action of the Tension. The distance from the axis to the line of action along this perpendicular is L*sin37 = moment arm for T. Draw perpendiculars from the axis to the lines of action of the the beam weight at the midpoint and the load at the right end and get moment arms L/2 and L, respectively. Summing the torques to zero give 0 = Lsin 37*T - (L/2)*Mg - L*mg, where M and m are the mass of beam and load, respectively . Solve for T. (b) Solve for F_{h}, the horizontal component of force on beam from pivot. F_{h} = T*cos37 Solve for F_{v}, the vertical component of force on beam from pivot: 0 = F_{v} + T*sin37 - mg - Mg, so F_{v} = (m + M)*g - T*sin 37. NOTE THE 3-4-5 TRIANGLE GIVES ME THE ANGLE OF 37 DEGREES. NOTE ALSO COS THETA = 4/5 = 0.8 AMD SIN THETA = 3/5 = 0.6. |
18. Choose axis at the beam left end. Find T using the torque
equation. Draw a perpendicular from the axis to the line of action
of the Tension. The distance from the axis to the line of action
along this perpendicular is (13 m)*sin25 = moment arm for T. Draw
perpendiculars from the axis to the lines of action of the beam
weight a distance of 7.0 m from the axis and the load at the right
end and get moment arms (7 m)*cos 55 and (16 m)*cos 55
respectively. (a) Summing the torques to zero gives 0 = (13 m)*sin 25*T - (7 m)*cos 55 *Mg - (16 m)*cos 55 *mg, where M and m are the mass of beam and load, respectively . Solve for T. (b) Solve for F_{h}, the horizontal component of force on beam from pivot. F_{h} = T*cos30. Solve for F_{v}, the vertical component of force on beam from pivot: 0 = F_{v} - T*sin30 - mg - Mg, so F_{v} = (m + M)*g + T*sin 30. |
3**, 11**, 10** (ladder--see example 11.3**), 12**, 13**,
14**, 18**, 24** (tensile tress,
strain---try 25**), 35** ( bulk modulus), 36** ( shear stress) 51**, 50** . |
22. See example 11.5. |
33. See example 11.6 |
34. See figure 11.17 and example 11.7 |
47. This is basically #14, with a sign hanging
vertically from the beam. Thus, this problem repeats what I
did in class. Choose axis at the beam left end. Find T using the
torque equation. Draw a perpendicular from the axis to
the line of action of the Tension. The distance from the axis to
the line of action along this perpendicular is L*sin theta =
moment arm for T. Draw perpendiculars from the axis to the lines
of action of the beam weight at the midpoint and the sign
toward the right end to get moment arms 0.75 m and 0.90 m,
respectively. Note: The midpoint (center of mass) of the sign is
0.6 m to the left of the right end, and thus is 0.90 m from the
axis at the left end. Summing the torques to zero give 0 = Lsin
41.4*T - (0.75 m)*Mg - (0.90 m) *mg, where M and m are the mass of
beam and sign, respectively . Solve for T. (b) Solve for F_{h}, the horizontal component of force on beam from pivot. F_{h} = T*cos41.4. Solve for F_{v}, the vertical component of force on beam from pivot: 0 = F_{v} + T*sin41.1 - mg - Mg, so F_{v} = (m + M)*g - T*sin 41.4. |
59. Choose axis at the beam left end. Find T using the
torque equation. Draw a perpendicular from the axis to the line of
action of the Tension. The distance from the axis to the line of
action along this perpendicular is L*sin20 = moment arm for T.
Draw a perpendicular from the axis to the line of action of the
beam weight a distance of L/2 from the axis and get moment arm
(L/2)*cos 40. (a) Summing the torques to zero gives 0 = L*sin 20*T - (L/2)*cos 40 *Mg, where M is the beam mass. Solve for T. (b) Solve for F_{h} = f_{s}, the horizontal component of force on beam from pivot. 0 = pos - neg = T*cos 60 - f_{s} , since the friction force points left. Thus f_{s} = T*cos 60 . Solve for F_{v}, the vertical component of force on beam from pivot: 0 = F_{v} + T*sin 60 - Mg, so F_{v} = M*g - T*sin 60. Now f_{smax} = u_{s}* N = u_{s}* F_{v} = u_{s}* ( M*g - T*sin 60 ) . But f_{smax} = T*cos 60. Solve for u_{s} . |