What is the acceleration of the small crate with a tension small enough to make sure it doesn't slide
HOME > Physics > What is the acceleration of the small crate with a tension small enough to make sure it doesn't slide

# What is the acceleration of the small crate with a tension small enough to make sure it doesn't slide

[From: ] [author: ] [Date: 11-10-07] [Hit: ]
but that is obviously the wrong answer. How do I go about this?2)In the previous situation, what is the frictional force the lower crate exerts on the upper crate?-1) T = 445 = (18 + 95) * accacc of system mass = 3.938m/s2acceleration of the small crateanswerstatic friction force between m1 & m2= 18 * 9.......
Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 18 kg and the larger bottom crate has a mass of m2 = 95 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is μs = 0.82 and the coefficient of kinetic friction between the two crates is μk = 0.66. 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).

1)The rope is pulled with a tension T = 445 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate?
-- Using Newton's 2nd law (F=ma) I originally thought it would be the same acceleration as the bottom crate, but that is obviously the wrong answer. How do I go about this?

2)In the previous situation, what is the frictional force the lower crate exerts on the upper crate?

-
1) T = 445 = (18 + 95) * acc
acc of system mass = 3.938m/s2
acceleration of the small crate
static friction force between m1 & m2
= 18 * 9.81 * 0.82
= 144.8 N
kinetic friction force between m1 & m2
= 18 * 9.81 * 0.66
= 116.54 N
2)frictional force the lower crate exerts on the upper crate
= 18 * 3.938
=70.88 N

since frictional force the lower crate exerts on the upper crate < kinetic friction force between m1 & m2, m1 will not move.

-
(F – μm1g) = ( m1 + m2) a

a = (F – μm1g) / (m1 + m2) = (445 -0.66*18*9.8)/ (95+18) = 2.09 m/s^2
======================================

2)0.66*18*9.8 = 116.4 N
==================

---------------------------------------…
The top crate will not slide => both should have the same rate of change of velocity ( the same acceleration) .Other wise there will be relative displacement.

The applied force is F = 445 N.
12
keywords: the,sure,slide,with,is,tension,039,acceleration,What,enough,it,of,to,doesn,crate,small,make,What is the acceleration of the small crate with a tension small enough to make sure it doesn't slide
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .