# HELP: Force and Inertia

1. Nov 2, 2004

### buffgilville

1) Two men are carrying a uniform plank of mass 5.6 kg and length 2.3 meters. A box of mass 108.4 kg is placed a distance 0.9 meters from the first man. Find the force applied by the second man.

Here what I did:
first man = force 1, second man force 2
F1 + F2 - mg = 0
F1 = (1.4)(5.6)(9.81) = 76.9104
F2 = mg - F1 = ((108.4)(9.81)) - 76.9104 = 986.4936

But the correct answer is 443.58. What did I do wrong?

2) A hollow ball of mass 1.9 kg and radius 0.1 meters begins rolling a distance 1.1 meters down a plane inclined at an angle 26 to the horizontal. Find its speed at the end.

Here is what I did:
Potential energy = (1.9)(1.1sin26)(9.81) = 8.988
KE = (1/2) (I) (w)^2
I=(2/3)(1.9)(0.1)^2 = 0.0127
PE = KE so, 8.988 = 0.00635w^2 ----> w = 37.622
w=v/r ----> v = (37.622)(0.1) ----> v = 3.76

2. Nov 2, 2004

### Pyrrhus

1) Looks like a Equilibrium Problem.

$$\sum_{i=1}^{n} \vec{F}_{i} = 0$$

$$\sum_{i=1}^{n} \vec{\tau}_{i} = 0$$

Where's the weight of the box??, you didn't include it.

2)

Kinetic Energy must include the kinetic energy for the linear movement

$$K = \frac{1}{2}I_{cm} \omega^2 + \frac{1}{2}mv^{2}_{cm}$$

remember

$$v_{cm} = r \omega$$

$$K = \frac{1}{2}I_{cm} \omega^2 + \frac{1}{2}mr^2 \omega^2$$

Applying Conservation of Mechanical Energy

$$mgdsin\theta = \frac{1}{2}I_{cm} \omega^2 + \frac{1}{2}mr^2 \omega^2$$

Solving for $\omega$

$$mgdsin\theta = \omega^2(\frac{1}{2}I_{cm} + \frac{1}{2}mr^2)$$

$$\sqrt{\frac{mgdsin\theta}{\frac{1}{2}I_{cm} + \frac{1}{2}mr^2}} = \omega$$

and

$$r \sqrt{\frac{mgdsin\theta}{\frac{1}{2}I_{cm} + \frac{1}{2}mr^2}} = v_{cm}$$

Note: $I_{cm} = \frac{2}{3}mr^2$ (Hollow Spherical shell)

Last edited: Nov 2, 2004
3. Nov 2, 2004

### buffgilville

T1 = F1(0)
T2 = F2(2.3)
T3 = -Mg(0.9)
T4 = -mg(2.3)
(T is torque, M is mass of box, m is mass of plank)

sum of T = F1(0) + F2(2.3) - Mg(0.9) - mg(2.3) = 0
F2 = Mg(0.9) + mg(2.3) - F1(0)
F2(2.3) = (108.4*9.81*0.9) + (5.6*9.81*2.3) + 0
so, the Torque of F2 is 471.05

T = Fr
so, (471.05) = F2 (1.4) -----> F2 = 336.46 Newtons

but the right answer is 443.58

Last edited: Nov 3, 2004
4. Nov 3, 2004

### Pyrrhus

buffgilville, for equilibrium problems we assume the weight of the body is at its center of gravity, this is because we consider uniform density and uniform gravitational field over the object, so the center of mass and center of gravity are in the same location. For an uniform rod its center of mass (Center of gravity) is half its length. This means the weight of the planck will act on its center of gravity.

5. Nov 3, 2004

6. Nov 3, 2004

### Pyrrhus

Understand what you're doing, rather than merely pursuing a number.

7. Nov 3, 2004

### buffgilville

Cyclovenom, you said that the center of mass (center of gravity) for an uniform rod is half its length;so, the weight of the plank will act on its center of gravity. That means that the axis of rotation is at the center of mass and r is the distance from the axis of rotation.
So,
T1 = F1(1.15)
T2 = F2(1.15)
T3 = -Mg(0.25)
T4 = -mg(2.3)
(T is torque, M is mass of box, m is mass of plank)

sum of T = F1(1.15) + F2(1.15) - Mg(0.25) - mg(2.3) = 0
F2(1.15) = Mg(0.25) + mg(2.3) - F1(1.15)
F2 = -722.36

I'm still getting it wrong, I read the whole chapter on static equilibrium, but I still am not getting it. Can you please help me work out this problem? Maybe that would help me understand better. (I am learning this on my own because my professor hasn't gone over this yet.)

8. Nov 3, 2004

### Pyrrhus

Ok, Here's the problem solved.

I will only need

$$\sum_{i=1}^{n} \vec{\tau}_{i} = 0$$

info:
$$d = 0.9 m$$
$$L = 2.3 m$$
$$m_{box}=108.4kg$$
$$m_{planck}=5.6kg$$

I will apply the torque condition where you applied it. (on the extreme we are not finding) The lever arm for the weight of the planck it's half the length of the planck, because that's where it acts (It's not necessarily the half, it depends on the origin for the torque sum, but the weight of the object acts on its center of gravity which varies its location depending on the body)

$$F_{man}L - m_{box}gd - m_{planck}g \frac{L}{2} = 0$$

$$F_{man} = \frac{m_{box}gd + m_{planck}g \frac{L}{2}}{L}$$

$$F_{man} \approx 443.58 N$$

Last edited: Nov 3, 2004
9. Nov 3, 2004

### buffgilville

Okay, I think I get it know. Thanks a BILLION Cyclovenom!!!