A block of mass m 1 on a rough, horizontal surface is connected to a ball of mas

Question

A block of mass m1 on a rough, horizontal surface is connected to a ball of mass m2 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 µk. Consider the system with m1 = 0.40 kg and m2 = 0.51 kg. The coefficient of static friction between the block and the surface is 0.45. The angle ? of the force F is equal to 28.0°. 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 block of mass m1 on a rough, horizontal surface is connected to a ball of mass m2 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 µk. Consider the system with m1 = 0.40 kg and m2 = 0.51 kg. The coefficient of static friction between the block and the surface is 0.45. The angle ? of the force F is equal to 28.0°. 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.

Explanation / Answer

Apply newton's law for the both blocks For the block(1): Apply Newton's law along horizontal direction Apply Newton's law along horizontal direction             Fcos? - fk - T = m1a       ---------(1) Apply Newton's law along vertical direction            n + F sin? - m1*g = 0                            n = m1*g - F sin?   The frictional force fk = µk(n) fk = µk (m1*g - F sin?)       -----------(2) From equtions (1) and (2) we get         Fcos? - µk (m1*g - F sin?) - T = m1a     Fcos? - µk *m1*g + µk*F sin? - T = m1a    Fcos? + µk*F sin? = m1a + µk *m1*g + T                           F = (m1a + µk *m1*g + T) / (cos? + µk sin? ) For ball         T - m2*g = m2*a          T = m2*g + m2*a Then we get         F = (m1a + µk *m1*g + m2*g + m2*a) / (cos? + µk sin? )    ---------(3) ---------------------------------------------------------------------------------- The masses are m1 = 0.40 kg , m2 = 0.51 kg The coefficient of static friction µs = 0.45 if the block is at rest, acceleration is zero. From the equation (3)         F = ( µs *m1*g + m2*g ) / (cos? + µs sin? )            F = ( 0.45 *0.40kg*9.8m/s^2 + 0.51kg*9.8m/s^2 ) / (cos28 + 0.45 sin28 )            F = 6.18 N This is the minimum force needed to keep the system is in equillibrium.                                                                                          Apply Newton's law along vertical direction            n + F sin? - m1*g = 0                            n = m1*g - F sin?   The frictional force fk = µk(n) fk = µk (m1*g - F sin?)       -----------(2) From equtions (1) and (2) we get         Fcos? - µk (m1*g - F sin?) - T = m1a     Fcos? - µk *m1*g + µk*F sin? - T = m1a    Fcos? + µk*F sin? = m1a + µk *m1*g + T                           F = (m1a + µk *m1*g + T) / (cos? + µk sin? ) For ball         T - m2*g = m2*a          T = m2*g + m2*a Then we get         F = (m1a + µk *m1*g + m2*g + m2*a) / (cos? + µk sin? )    ---------(3) ---------------------------------------------------------------------------------- The masses are m1 = 0.40 kg , m2 = 0.51 kg The coefficient of static friction µs = 0.45 if the block is at rest, acceleration is zero. From the equation (3)         F = ( µs *m1*g + m2*g ) / (cos? + µs sin? )            F = ( 0.45 *0.40kg*9.8m/s^2 + 0.51kg*9.8m/s^2 ) / (cos28 + 0.45 sin28 )            F = 6.18 N This is the minimum force needed to keep the system is in equillibrium.                      Fcos? + µk*F sin? = m1a + µk *m1*g + T                           F = (m1a + µk *m1*g + T) / (cos? + µk sin? ) For ball         T - m2*g = m2*a          T = m2*g + m2*a Then we get         F = (m1a + µk *m1*g + m2*g + m2*a) / (cos? + µk sin? )    ---------(3) ---------------------------------------------------------------------------------- The masses are m1 = 0.40 kg , m2 = 0.51 kg The coefficient of static friction µs = 0.45 if the block is at rest, acceleration is zero. From the equation (3)         F = ( µs *m1*g + m2*g ) / (cos? + µs sin? )            F = ( 0.45 *0.40kg*9.8m/s^2 + 0.51kg*9.8m/s^2 ) / (cos28 + 0.45 sin28 )            F = 6.18 N This is the minimum force needed to keep the system is in equillibrium.         F = ( µs *m1*g + m2*g ) / (cos? + µs sin? )            F = ( 0.45 *0.40kg*9.8m/s^2 + 0.51kg*9.8m/s^2 ) / (cos28 + 0.45 sin28 )            F = 6.18 N This is the minimum force needed to keep the system is in equillibrium.         F = 6.18 N This is the minimum force needed to keep the system is in equillibrium.                                                                             

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.