ENSAYO DE CONTENIDO DE AIRE EN MORTEROS NORMA: NTE INE
5. DISEÑO DE LAS MEZCLAS DE PRUEBA
5.5. Mezclas de prueba (Alternativas de mezcla).
tant, because it provides the quantitative relationship between force and acceleration. The
examples in this section serve as a review of the essential features of this relationship.
m1 m2 m1 (a) (b) m2 Pushing force Pushing force
Background: This question deals with net force, acceleration, and Newton’s second and third laws of motion.
For similar questions (including calculational counterparts), consult Self-Assessment Test 4.3. The test is described next.
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Concepts &Calculations Example 20
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Velocity, Acceleration, and
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Newton’s Second Law of Motion
Figure 4.38 shows two forces, 1 3000 N and 2 5000 N, acting on a spacecraft, where the plus signs indicate that the forces are directed along the x axis. A third force 3 also acts on the spacecraft but is not shown in the drawing. The craft is moving with a constant velocity of 850 m/s. Find the magnitude and direction of 3.
Concept Questions and Answers
Suppose the spacecraft were stationary. What would be the direction of 3?Answer If the spacecraft were stationary, its acceleration would be zero. According to Newton’s second law, the acceleration of an object is proportional to the net force acting on it. Thus, the net force must also be zero. But the net force is the vector sum of the three forces in this case. Therefore, the force 3 must have a direction such that it balances to zero the forces 1and 2. Since 1and 2point along the x axis in Figure 4.38, 3must then point along the x axis.
When the spacecraft is moving at a constant velocity of 850 m/s, what is the direction of 3?
Answer Since the velocity is constant, the acceleration is still zero. As a result, every- thing we said in the stationary case applies again here. The net force is zero, and the force
3must point along the x axis in Figure 4.38.
Solution
Since the velocity is constant, the acceleration is zero. The net force must also be zero, so thatSolving for F3yields
The minus sign in the answer means that 3points opposite to the sum of 1and 2, or along the x axis in Figure 4.38. The force 3has a magnitude of 8000 N, which is the magnitude of the sum of the forces 1and 2. The answer is independent of the velocity of the space- craft, as long as that velocity remains constant.
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Concepts &Calculations Example 21
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The Importance of Mass
On earth a block has a weight of 88 N. This block is sliding on a horizontal surface on the moon, where the acceleration due to gravity is 1.60 m /s2. As Figure 4.39a shows, the block is
being pulled by a horizontal rope in which the tension is T 24 N. The coefficient of kinetic friction between the block and the surface is k 0.20. Determine the acceleration of the
block.
Concept Questions and Answers
Which of Newton’s laws of motion provides a way to de- termine the acceleration of the block?Answer Newton’s second law allows us to calculate the acceleration as ax Fx/m, where Fx is the net force acting in the horizontal direction and m is the mass of the block.
This problem deals with a situation on the moon, but the block’s mass on the moon is not given. Instead, the block’s earth-weight is given. Why can the earth-weight be used to obtain a value for the block’s mass that applies on the moon?
Answer Since the block’s earth-weight Wearthis related to the block’s mass according to
Wearth mgearth, we can use Wearth 88 N and gearth 9.80 m/s2to obtain m. But mass is
an intrinsic property of the block and does not depend on whether it is on the earth or on the moon. Therefore, the value obtained for m applies on the moon as well as on the earth.
Does the net force Fxequal the tension T ?
Answer No. The net force Fx is the vector sum of all the external forces acting in the horizontal direction. It includes the kinetic frictional force fkas well as the tension T.
F B F B F B F B FB F B 8000 N F3 (F1 F2) (3000 N 5000 N) Fx F1 F2 F3 0 FB F B FB F B F B F B F B F B F B F B F B F B F B F2 F1 +x B B
Figure 4.38 Two horizontal forces, 1
and 2, act on the spacecraft. A third
force F3also acts but is not shown. B F B F B T ax (a)
(b) Free-body diagram for the block +x +y FN fk T mg moon B B
Figure 4.39 (a) A block is sliding on a horizontal surface on the moon. The tension in the rope is . (b) The free- body diagram for the block, including a kinetic frictional force fk.
B
T
B
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Solution
Figure 4.39b shows the free-body diagram for the block. The net force along the x axis is Fx T fk, where T is the magnitude of the tension in the rope and fkis the mag-nitude of the kinetic frictional force. According to Equation 4.8, fkis related to the magnitude
FNof the normal force by fkkFN, where kis the coefficient of kinetic friction. The ac-
celeration axof the block is given by Newton’s second law as
We can obtain an expression for FNby noting that the block does not move in the y direction,
so ay 0 m/s2. Therefore, the net force F
yalong the y direction must also be zero. An ex- amination of the free-body diagram reveals that Fy FN mgmoon 0, so that FN
mgmoon. The acceleration in the x direction becomes
Using the earth-weight of the block to determine its mass, we find
The acceleration of the block is, then,