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iurod
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Homework Statement
Two crates, of mass 75 kg and 110 kg are in contact and at rest on a horizontal surface. A 620N force is exerted on the 75 kg crate. If the coefficient of kinetic friction is 0.15 calculate:
a) the acceleration of the system
b)the force that each crate exerts on the other
* I have attached a word document with the picture my book gave me, hoping that it helps.
Homework Equations
f=ma
Friction= (coefficient of kinetic friction)(mg)
The Attempt at a Solution
a) the acceleration of the system
for the Free Body Diagram I chose to make one big block with mass= 185kg
f=ma
620N - Friction = (185 kg)a
Friction= (0.15)(185x9.8) = 271.95
620 - 271.95 = (185 kg)a
a= 1.9 m/s2
b. the force that each crate exerts on the other
This is where I'm having trouble, I could find the formula nor did our professor give us the formula for Contact force. So I tried to think it out and got this
Contact Force = Force - Friction - ma
taking the 75kg box first:
620 - Friction -(75 kg)(1.9 m/s2)
Friction = (0.15)(75x9.8) = 110.25
Contact force = 620 - 110.25 - (75)(1.9)
Contact force = 367.25 I rounded to 3.7x102N
Now for the 110 kg box:
620 - Friction - (110)(1.9)
Friction = (0.15)(110x9.8) = 161.7
Contact force = 620 - 161.7 - (110)(1.9)
Contact force = 249.3 I rounded to 2.5x102
Thanks!