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Homework Help: Rotational Equilibrium and Dynamics Help

  1. May 17, 2005 #1
    the following four questions were from my test that i recentally took and i miss the day that we went over them and didn;t get the answers and haven;t had to time to check with my teacher about them, so i need some help on what the answers would be. thanks.

    1. If the torque required to loosen a nut is 40.0 mN, what miminum force must be exerted by a mechanic to the end of a 30.0 cm wrench to remove the nut?

    2. A uniform bridge 20.0 m long and weighting 4.00 x 10^5 N is supported by pillars located 3.00 m from each end. if a 1.96 x 10^4 N car is parked 8.00 m from one end of the bridge, how much force does each pillar exert.

    3. A person lifts a 950 N box by pushing it up a incline. If the person exerts a force of 350 N along the incline, what is the mechanical advantage of the incline?

    4. A 35 kg bowling ball with a radius of 13 cm starts from rest at the top of an incline 3.5 m in height. Find the translational speed of the bowling ball after it has rolled to the bottom of the incline. (Assume the ball is a uniform solid sphere.)
  2. jcsd
  3. May 18, 2005 #2


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    1. is simply a "solve for the answer" type of question. What you gotta be able to do though is translate what you are being told in words into actual mathematical formulae. Here, this would look like:

    We know the formula for torque is

    [tex]\vec{\tau} = \vec{r}\times \vec{F} = rFsin\theta[/tex]

    For a given torque and a given distance, for which angle is the force required minimal? When [itex]sin\theta[/itex] is maximal, indeed! That is, when [itex]sin\theta = 1 \Leftrightarrow \theta = \pi/2[/itex]. So when the minimal force occurs, the equation of torque is simply

    [tex]\tau = rF[/tex]

    Now that this theoretical stuff is out of the way, on to the computation: We are told a torque of 40 Nm is required to loosen the nut. We want to find the force F that produces a torque of 40 Nm if r = 30.0cm = 0.300m. So we want F such that

    [tex]40 Nm = 0.300m\cdot F \Leftrightarrow F = 40Nm/0.300m = 133.3N[/tex]

    2. This one requires more thinking. First of all we draw the free body diagram, we give ourselves a conventionnal coordinate system (positive y points at the sky) and label all the forces acting on the bridge. It is obvious that the bridge is in both translationnal and rotational equilibrium. This means that both the sum of forces and sum of torques --about any axis-- acting on the bridge are zero. Let's write that down mathematically:

    [tex]\sum F = 0 \ \ \ \ \ \ \ \sum \tau = 0[/tex]

    What are the forces? Well first there's the force that the car exerts on the bridge through gravity. This force is of -1.96 x 10^4 N. There's also the force of gravity from the earth acting on the bridge. This one is of -4.00 x 10^5 N. And finally, there are the two normal forces from the pilars. We will label those [itex]N_1[/itex] and [itex]N_2[/itex]. So we have the equation

    [tex]\sum F = -1.96 \cdot 10^4N -4.00 \cdot 10^5N + N_1 + N_2 = 0[/tex]

    What are the torques? Let's compute them about the axis passing through the center of mass (CM) of the bridge (its geometrical center). Well they're simply the product of the forces enumerated above with the distances from the CM at which they act. This yields

    [tex]\sum \tau = 1.96 \cdot 10^4N \cdot 2m + 0 + N_1\cdot 7m - N_2\cdot 7m = 0[/tex]

    The 0 in there represents the torque made by the force of gravity. It is zero because the torque of every particles on the left of the CM is positive, while the torque of every particle on the right is positive. So there being as much as much mass at the right as at the left, by symetry, the torques from the particles on the left cancel each the torques from the particles on the right.

    So in the end, what we are left with are two equations in two unknown, namely N_1 and N_2. Solve for them.

    3. The box weight -950N. So usually, in order to lift it a distance h in the air, one would have to exert a force of +950N. But with the incline, the guy managed to pull it off deploying a force of only 350N, the rest being suplied by the normal force of the incline. I'm not sure if this is the answer your teacher was looking for, or if he gave a special meaning to "mechanical advantage".

    4. I think this is a trick question, in the sense that the radius of the ball is not required in solving this problem. By conservation of energy, it is imediate:

    [tex]\Delta T = \Delta U[/itex]

    [tex]\Delta U = mg(h_f - h_i) = 35kg\cdot 9.8 ms^{-2}\cdot 3.5m = 1200.5J[/tex]

    [tex]\Delta T = \frac{1}{2}m(v_f^2 - v_i^2) = \frac{1}{2}\cdot 35kg\cdot v_f^2[/tex]

    Solve for v_f².
  4. May 18, 2005 #3
    The mechanical advantage of a machine is the ratio of the load to the effort.

    When a machine outputs more force than is put in, the machine is said to have mechanical advantage.
  5. May 18, 2005 #4


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    Oh ok, so the answer the teacher was looking for was really a number. So.. 950/350 ?
  6. May 18, 2005 #5
    so what do you get for N_1 and N_2. I understand the other parts but am confused on how to get the answer for n_1 and n_2. I believe for #3 he is looking for a word answer and i know that 950/350 is not correct from asking kids in class about it.
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