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Two crates, of mass m1 = 60 kg and m2 = 110 kg, are in contact and at rest on a horizontal surface (Fig. 4-54). A 620 N force is exerted on the 60 kg crate. The coefficient of kinetic friction is 0.15.

a) Calculate the acceleration of the system.
=............ m/s^2 (to the right)

(b) Calculate the force that each crate exerts on the other.
=.......N

(c) Repeat with the crates reversed.
acceleration =......m/s2
crate force=........ N

2007-09-27 06:40:01 · 1 answers · asked by Nikita 1 in Science & Mathematics Physics

1 answers

Are the crates stacked one on the other?

Start with the 60 on top of the 110

a FBD of the 60 shows
620-f=60*a
where f is the force of friction between the two crates

a FBD of the 110 shows

f-F=110*a

where F is the friction at the horizontal surface

If we assume that the crates do not slip between each other, then
F=170*g*0.15
using g=9.81
F= 250 N

now solve for f and a

f=(110*620+60*250)/170

f=489 N

j

2007-09-27 07:03:01 · answer #1 · answered by odu83 7 · 0 0

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