(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant equations

f=ma

Friction= (coefficient of kinetic friction)(mg)

3. 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/s^{2}

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/s^{2})

Friction = (0.15)(75x9.8) = 110.25

Contact force = 620 - 110.25 - (75)(1.9)

Contact force = 367.25 I rounded to 3.7x10^{2}N

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 2.5x10^{2}

Thanks!

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Newtons Law and Friction Question (along with a contact force question)

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