Two wooden crates rest on top of one another. The smaller top crate has a mass o
ID: 1531543 • Letter: T
Question
Two wooden crates rest on top of one another. The smaller top crate has a mass of m_1 = 19 kg and the larger bottom crate has a mass of m_2 = 87 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is mu_s = 0.87 and the coefficient of kinetic friction between the two crates is mu_k = 0.69. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem). The rope is pulled with a tension T = 335 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate? In the previous situation, what is the frictional force the lower crate exerts on the upper crate? What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide? The tension is increased in the rope to 1221 N causing the boxes to accelerate faster and the top box to begin sliding. W What is the acceleration of the upper crate? As the upper crate slides, what is the acceleration of the lower crate?Explanation / Answer
1 When the top crate does not slide, the two crates will have theSAME acceleration to the right. Use Newton’s second law,
T = (m1 + m2) a
(There is NO friction between the crate and the floor).Or
335 = (19 + 87) a
a = 3.16 m/s2
2. F = m1a = 19* 3.16=60.04N
3. Fmax = s FN (FN = Normal force exerted on the upper crate by the lower crate)
Fmax = s m1g
The maximum acceleration is
amax =Fmax/m1 = s g = 0.87 * 9.8 = 8.526 m/s2
The maximum tension is Tmax = (m1 + m2) amax = (19 + 87) *3.16 = 334.96N
4.Fk = k FN (FN = Normal force exerted on the upper crate by the lower crate)
Fk = k m1g
The acceleration is a1=Fk/m1= k g = 0.69 *9.8=6.762 m/s2
5. Fnet = 1221 – Fk = 1221 - k m1g = 1221 – 0.69 *19 *9.8=1092.522N
The acceleration of the lower crate is given by
a2 =Fnet /m2=12.56m/s^2
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