# AP Physics Static Friction and Compression Question?

A 4.0-m long steel beam with a cross-sectional area of 1.0 × 10−2 m^2 and a Young’s modulus of 2.0 × 10^11 N/m^2 is wedged horizontally between two vertical walls. In order to wedge the beam, it is compressed by 0.020 mm. If the coefficient of static friction between the beam and the walls is 0.70 the maximum mass (including its own) it can bear without slipping is:

A. 0

B. 3.6 kg

C. 36 kg

D. 71 kg

E. 710 kg

I set friction equal to the downward force of weight and solved to get that weight (W)= 2μ(EA(ΔL/L)), however when I plug the given values into this formula, I don't get the correct answer of 710 kg. Please help, thank you!

### 2 Answers

- GFLv 65 years agoFavorite Answer
OK Let me try.

If you draw a free body diagram of the steel beam wedged across two vertical walls. You will see that there is a force of Friction on both LHS of the beam and RHS of the beam. It has static friction = 0.7 and this Force = mu * Normal Force. So we have 2*Force Upward which must support mass * gravity downward. That is the balance.

Normal Force is the force of compression of the steel beam. We know it is compressed by 0.020 mm = 0.020 * 10^-3 meters. I hope this helps

dl = F*Lo /( AE) ---> Fcompression = dl * AE / Lo (steel beam compression force) = 10,000 N This force must be balance by the Normal force on each vertical wall. So the Normal force on LHS + Normal force on RHS = Force of compression. So each Normal force is 5000 N because 5kN + 5kN = 10kN and 10kN is the compression force.

2 * mu * Normal Force = mg -----> m = 2 * mu * Normal / g = 2 * 0.7 * 5000 / 9.81 = 713 kg

I think this is correct

- civil_av8rLv 75 years ago
Use the compression to find the force applied. Since it is wedged, this is essentially the normal force. Once you have that then simply use friction = u*N. I'll help solve this later.

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Stress = e*E

Stress = F/A

F/A = e*E

F = A*e*E

Li = 4 m

Lf = 4 m - 0.020 mm = 4 m - .00002 m = 3.99998 m

e = (Li - Lf)/Li = .00002 m / 4 m = 0.000005

F = 1.0*10^-2 m^2 * 0.000005 * 2.0*10^11 N/m^2 = 10000 N

Like I said earlier, friction = u*F

Sum of the vertical forces = 0 = m*g - friction

m = friction / g = u*F/g

Plug in numbers

m = 0.70 * 10000 N / 9.81 m/s^2 = 710 kg <----

Hope the logic made sense so you can understand the concepts.