ANSWERS

Quiz 9---CH. 9: 1, 2, 4, 11, 15, 16, 18, 20, 21, 63

Turn in: 2, 15, 18, 20, 21

2.| torque |= (3.0 m)* weight, where person's weight = mg. See formula 8-10a): |torque| = (rsinθ)F. Here θ = 90 degrees and F = mg. Note: sin90 = 1. Review section 8.4 for definitions of torque.
Note that above we used the absolute value of the torque. The women tends to rotate the diving board in the clockwise direction which is considered the negative direction; so the torque would be  negative if clockwise was considered negative. What is important is the absolute value (magnitudes). We will subtract magnitudes in the problems below to express the equilibrium condition. 

15.  See example 9.4 and especially example 9.5.  Here is how it goes:

You have two unknowns F_A and F_B. Thus , you need two equations. The first obvious equation is one of translational equilibrium,

(I) The sum of the forces in the y-direction must add to zero. 
Thus: 0 = pos - neg = F_A + F_B  - 4300 - 3100 - 2200 - mg.  (I)
The second equation is one of rotational equilibrium. Choose the left end of the beam as the axis of rotation.
The sum of the torques = 0 = pos - neg
0 = (10 m)*F_B  - (2.0 m)*(4300 N) - (6 m)*(3100 N) - (9 m)*(2200 N) - (5.0 m)*mg. (II).
Note that in the torque equation, the force at the left end of the beam does not appear because it exerts zero torque about that axis. 

Solve (I) and (II) for the two unknown forces. Note that weight of the beam is mg and acts at the center of mass , which is at the geometrical center of the beam according to section 7.8  

(I) The sum of the forces in the y-direction must add to zero. 
Thus: 0 = pos - neg = F_V + Tsin40 - mg.  (I)
(II) The sum of the forces in the x-direction must add to zero. 
Thus: 0 = pos - neg = F_H  - Tcos40 .  (II)
The third equation is one of rotational equilibrium. Choose the left end of the beam as the axis of rotation.
The sum of the torques = 0 = pos - neg
0 = L*Tsin40 - (L/2)*mg  (III).

Note that you can immediately solve for T in equation (III). Note: L cancels. Plug your value of T into equation (I) and (II) to get F_H   ands F_V   . 

Solve (I) , (II) and (III) for the three  unknowns. 

Note that weight of the beam is mg and acts at the center of mass, which is at the geometrical center of the beam according to section 7.8   

Note to get the magnitude of the net force of the wall on the beam, just compute that magnitude from the two force components   F_H   ands F_V   using the  Pythagorean Theorem. You can also get the angle that the net force makes with the horizontal . Use the trig. result that angle = Tan^-1( F_V  / F_H)

Here is how it goes:
You have three unknowns T , F_V  and  F_H .  These are,  respectively, the magnitude T of the tension force, the vertical component F_V  of the force of the wall on the beam, and the horizontal  component F_H  of the force of the wall on the beam.  Thus, you need three equations. The first two obvious equations are ones of translational equilibrium,

(I) The sum of the forces in the y-direction must add to zero. 
Thus: 0 = pos - neg 
0 = F_V + Tsin35 - mg - Mg, where mg = 155 (N) and Mg = 245 (N).  (I)
(II) The sum of the forces in the x-direction must add to zero. 
Thus: 0 = pos - neg = F_H  - Tcos35 .  (II)
The third equation is one of rotational equilibrium. Choose the left end of the beam as the axis of rotation.
The sum of the torques = 0 = pos - neg
0 = (1.35 m)*Tsin35 - (L/2)*mg - L*Mg  (III).
Note: L = 1.70 m.

Note that you can immediately solve for T in equation (III). Plug your value of T into equation (I) and (II) to get F_H   ands F_V     .

Solve (I) , (II) and (III) for the three  unknowns. 

Note that weight of the beam is mg and acts at the center of mass, which is at the geometrical center of the beam according to section 7.8.  The sign weight Mg acts at the end of the beam.  

Note to get the magnitude of the net force of the wall on the beam, just compute that magnitude from the two force components   F_H   ands F_V   using the  Pythagorean Theorem. You can also get the angle that the net force makes with the horizontal . Use the trig. result that angle = Tan^-1( F_V  / F_H)

You have three unknowns T , F_V  and  F_H .  These are,  respectively, the magnitude T of the tension force, the vertical component F_V  of the force of the wall on the beam, and the horizontal  component F_H  of the force of the wall on the beam.  Thus, you need three equations. The first two obvious equations are ones of translational equilibrium,

(I) The sum of the forces in the y-direction must add to zero. 
Thus: 0 = pos - neg 
0 = F_V + Tsin0 - mg - Mg, where m = 12.0 kg and Mg = 21.5 kg.  (I)
Note: sin0 = 0.
(II) The sum of the forces in the x-direction must add to zero. 
Thus: 0 = pos - neg = F_H  - Tcos0 .  (II)
Note: cos0 = 1.
The third equation is one of rotational equilibrium. Choose the left end of the beam as the axis of rotation.
The sum of the torques = 0 = pos - neg
0 = (3.8 m)*T - (L/2)*(cos37)*mg - Lcos37*Mg  (III).
Note: L = 7.50 m.

Note that you can immediately solve for T in equation (III). Plug your value of T into equation (I) and (II) to get F_H   ands F_V     .

Solve (I) , (II) and (III) for the three  unknowns. 

Note that weight of the beam is mg and acts at the center of mass, which is at the geometrical center of the beam according to section 7.8.  The traffic light weight Mg acts at the end of the beam.  

Note to get the magnitude of the net force of the wall on the beam, just compute that magnitude from the two force components   F_H   ands F_V   using the  Pythagorean Theorem. You can also get the angle that the net force makes with the horizontal . Use the trig. result that angle = Tan^-1( F_V  / F_H). 
  

 More hints later.