 ToC, Ch 8  Course Calendar
8.3 A 2kg ball is held in position by a
horizontal string and a string that makes an angle of 30° with
the vertical, as shown in the figure. Find the tension T in the
horizontal string.
(Since I am writing this all in bold to begin with, it is difficult or impossible to explicitly tell that this is a vector equationbut it is! And that's very important!)
But that really means
Apply the first condition of equilibrium to
each pulley.
First to the one one the left,
That gives two unknowns in one equation so we must seek additional information. Of course, that will come from looking at the pulley on the right,
The maximum external force F that can be exerted will be the force exerted when the friction forces are at their maxima,
Make careful free body diagrams of the forces
on each block.
First, for the 2kg block, we have
But that really means
With the normal force on the 2 kg block known, we can readily
calculate the friction force,
And that also means
Now, look at all the forces on the 7kg block,
The downward 20 Nforce on top of the block is the normal force exerted by the 2kg block. The 8N force on the top is the friction force exerted by the 2kg block. The tension is still 8 N just as it was at the other end of the string which is attached to the 2kg block. The forces labeled F_{N} and F_{f} are the normal force and friction force exerted at the bottom on this 7kg block, of course. We used the same symbols to represent different forces earlier when we looked at the 2kg block. Now we are ready to again apply the first condition of equilibrium,
But that really means








With the normal force on the 7kg block known, we can readily calculate the friction force,
= r F sin
= (0.30 m) (150 N) (1)
= 45 mN
As always, a free body diagram is
essential.
From the first condition of equilibrium, we have
But that really means








More information is needed to solve for T and F_{x}. We can get that information from the second condition of equilibrium. Before we start to calculate torques, we must decide on the reference point about which we will calculate those torques. If we choose the lower, left end of the pole, there are two forces, F_{x} and F_{y}, that will have zero torque. That will reduce the number of terms in all of our equations. Therefore, that is a good choice for the origin or reference point or axis of rotation.
F_{x}:
= 0 (since r = 0 in
= r F sin
)
F_{y}:
= 0 (since r = 0 in
= r F sin
)
T: _{ccw}
= (4 m)(T)(sin 30°) = (4 m) T
(0.50)
=
(2
m) T
600 N: _{cw}
= (6 m)(600 N)(sin 60°) = (6 m)(600
N)(0.866)
=
3118
mN
300 N: _{cw} = (3 m)(300 N)(sin 60°) = (3 m)(300 N)(0.866) = 779 mN
_{ccw}
=
_{cw}_{
}(2 m) T = 3118 mN + 779
mN
(2 m) T = 3897 N
T = (3897/2) N
T = 1948.5 N
Therefore,
F_{x} = 1948.5 N
And we already knew
As always, a free body diagram or a force
diagram is
essential.
From the first condition of equilibrium, we have
But that really means








More information is needed to solve for T and F_{x}. We can get that information from the second condition of equilibrium. Before we start to calculate torques, we must decide on the reference point about which we will calculate those torques. If we choose the lower, left end of the pole, there are two forces, F_{x} and F_{y}, that will have zero torque. That will reduce the number of terms in all of our equations. Therefore, that is a good choice for the origin or reference point or axis of rotation.
F_{x}:
= 0 (since r = 0 in
= r F sin
)
F_{y}:
= 0 (since r = 0 in
= r F sin
)
T: _{ccw}
= (4 m)(T)(sin 30°) = (4 m) T
(0.50)
=
(2
m) T
600 N: _{cw}
= (6 m)(600 N)(sin 60°) = (6 m)(600
N)(0.866)
=
3118
mN
_{ccw}
=
_{cw}_{
}(2
m) T =
3118
mN
T =
1559
N
Therefore,
F_{x} = 1559 N
And we already knew
Think of the square as being made of two rectangles.
The center of gravity of each rectangle is at the geometric center of that rectangle and the mass (or weight) located at that position is proportional to the area of that rectangle.
Notice that the positions are all measured from the lower left corner of the carpenter's square. Now we can use the defintions of center of gravity, Equations 8.10 and 8.11, to locate the center of gravity of the entire object,
Using the first condition of equilibrium, we have
F_{wall} = F_{f}
and
F_{N} = 1200 N
To solve for F_{wall} and, thus, for F_{f}, we must use the second condition of equilibrium. In calculating the torques, let us calculate torques about the foot of the ladder. That choice means two of the forces, F_{N} and F_{f }, will provide zero torque and that reduces the number of terms in our equations. Be very careful of the angles!
F_{wall}: _{cw} = (6.0 m)(F_{wall})(sin 60°) = (6.0 m)(F_{wall})(0.866)_{cw} = (5.2 m) F_{wall}800 N: _{ccw} = (6.0 m)(800 N)(sin 30°) = 2400 mN
400 N: _{ccw} = (2.0 m)(400 N)(sin 30°) = 400 mNF_{wall}: = 0
F_{f}: = 0
Call the components of the force exerted by the pin F_{x} and F_{y}. The angle between the guy wire and the horizontal is
Start with the first condition of equilibrium,
F_{x}
= 0.80 T
F_{y}
=
F_{y}
+ T sin 37°  600 N = 0
F_{y}
+ 0.60 T = 600 N
This leaves us with two equations but three unknowns, so some additional information is required. Of course, we can get that by applying the second condition of equilibrium. We will calculate torques about the pin. This means the torque caused by the force of the pin is zero. Be careful of the torque for the 600N weight.
T: _{ccw} = (2.0 m)(T)(sin 37°) =(2.0 m)(T)(0.60) =(1.2 m)T
F_{x}: = 0
F_{y}: = 0
W: _{cw} = (1.0 m)(600 N) = 600 mN
From the first condition of equilibrium, we have
Calculate torques about the position of the child,
W: = 0
F_{top}: _{ccw} = (18 cm)F_{top}
F_{bottom}: _{cw} = (45 cm) F_{bottom}
Now apply the second condition of equilibrium,
Ooops, I forgot to give you values for
w_{1} and
w_{2};
w_{3} is the weight of
294 N (the 30kg mass that is being lifted). Read problem 8.73 for
those values;
w_{1}
= 480 N and w_{w} = 60
N.
From the first condition of equilibrium we have
and
We now have two equations with three unknowns so we must get additional information from the second condition of equilibrium. We will calculate torques about the hip; that means the torque exerted by the compressive force C will be zero.
C: = 0
T: _{ccw} = (48 cm) T (sin 12°) = (48 cm) T (0.208)= (9.98 cm) Tw_{1}: _{cw} = (36 cm)(480 N) = 17 280 cmN
w_{2}: _{cw} = (72 cm)(60 N) = 4 320 cmN
w_{3}: _{cw} = (48 cm)(294 N) = 14 112 cmN
The negative sign here simply means that while we assumed that C_{y} pointed down, C_{y} actually points up.
 ToC, Ch 8  Course Calendar
(c) 2000, Doug Davis; all rights reserved.