Now consider a surface with friction: If the coefficient of static friction is .

Question

Now consider a surface with friction: If the coefficient of static friction is .30 and the coefficient of kinetic friction is .20 what is the maximum incline of the surface for the applied force of 42 N to be able to move the blocks?

Assume that three blocks shown above move on a frictionless surface and that 42N force acts as shown on block box #3. The surface is now incline. What must be the angle of the inclination if the 42N force moves the system with constant speed. The answer is 45.7 degrees. Now consider a surface with friction: If the coefficient of static friction is .30 and the coefficient of kinetic friction is .20 what is the maximum incline of the surface for the applied force of 42 N to be able to move the blocks?

Explanation / Answer

Let's take a Free body doagram of block 3.

42N force is pulling it up, Tension T is pulling it down. Also, mgcos will be pulling it down. Im taking g as 10 for simplicity.

Thus, by force balance since at constant speed there is no net force, 42 = T + 30cos ......Eq1

Now, take block 1. Tension T pulls it up, 10cos pulls it down and there is also the Reaction force R from block 2.

Thus, T = 10cos+R ....eq2

Now for block 3, the reaction R pulls it up and its weight tries to bring it down.

Thus, R = 20cos.....Eq3

substitute equation 3 in eq2.

T = 30cos.

Substitue this result in eq 1 and you get 42 = 60cos.

Thus = cos-1 (42/60).

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