Example 5.13 Acceleration of Two Connected Objects When Friction Is Present A bl
ID: 1659599 • Letter: E
Question
Example 5.13 Acceleration of Two Connected Objects When Friction Is Present
A block of mass m2 on a rough, horizontal surface is connected to a ball of mass m1 by a lightweight cord over a lightweight, frictionless pulley as shown in figure (a). A force of magnitude F at an angle with the horizontal is applied to the block as shown, and the block slides to the right. The coefficient of kinetic friction between the block and surface is k. Determine the magnitude of the acceleration of the two objects.
SOLVE IT
Apply the particle under a net force model to the block in the horizontal direction:
(1)
Fx = F cos fk T = m2ax = m2a
Because the block moves only horizontally, apply the particle in equilibrium model to the block in the vertical direction:
(2)
Fy = n + F sin m2g = 0
Apply the particle under a net force model to the ball in the vertical direction:
(3)
Fy = T m1g = m1ay = m1a
Solve Equation (2) for n:
n = m2g F sin
Substitute n into fk = kn:
(4) fk = k(m2g F sin )
Substitute Equation (4) and the value of Tfrom Equation (3) into Equation (1):
F cos k(m2g F sin ) m1(a + g) = m2a
Solve for a (Use the following as necessary: k, m1, m2, g, F, and ):
(5) a =
F(cos()+ksin())(m1+km2)gm1+m2
Finalize The acceleration of the block can be either to the right or to the left depending on the sign of the numerator in Equation (5). If the motion is to the left, we must reverse the sign of fk in Equation (1) because the force of kinetic friction must oppose the motion of the block relative to the surface. In this case, the value of a is the same as in Equation (5), with the two plus signs in the numerator changed to minus signs.
MASTER IT
Consider the system described in the example with m1 = 0.51 kg and m2 = 0.40 kg. The coefficient of static friction between the block and the surface is 0.45. The angle of the force F is equal to 25.0°. If F = 0, you can easily show that the block will accelerate to the left since the maximum static friction force is not sufficient to keep the block at rest. If F is sufficiently large, it is clear that the block will accelerate to the right. Find the range of F that allows the system to remain at rest.
(a) The external force applied as shown can cause the block to accelerate to the right. (b, c) Diagrams showing the forces on the two objects, assuming the block accelerates to the right and the ball accelerates upward.Explanation / Answer
for minimum
along horizontal
Fx = F*costheta + fk - T = 0 ........(1)
along vertical
for m2
Fy = n + F*sintheta - m2*g = 0
n = m2*g - F*sintheta
frictional force fk = uk*n = uk*(m2*g - F*sintheta)........(2)
for m1
m1*g - T = 0
T = m1*g .........(3)
therefore
substitute 2 , 3 in 1
F*costheta + uk*(m2*g - F*sintheta) - m1*g = 0
F*cos25 + 0.45*((0.4*9.8)-(F*sin25)) - (0.51*9.8) = 0
Fmin = 4.52 N <<<------ANSWER
========================================
for maximum
along horizontal
Fx = F*costheta - fk - T = 0 ........(1)
along vertical
for m2
Fy = n + F*sintheta - m2*g = 0
n = m2*g - F*sintheta
frictional force fk = uk*n = uk*(m2*g - F*sintheta)........(2)
for m1
m1*g - T = 0
T = m1*g .........(3)
therefore
substitute 2 , 3 in 1
F*costheta - uk*(m2*g - F*sintheta) - m1*g = 0
F*cos25 - 0.45*((0.4*9.8)-(F*sin25)) - (0.51*9.8) = 0
Fmax = 6.17 N <<<------ANSWER
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