Phys Static Friction Ramp Problem

[PhysicsDerp]

Homework Statement

The problem reads : You are riding your motorcycle one day down a wet street that slopes downward at an angle of 20 ∘ below the horizontal. As you start to ride down the hill, you notice a construction crew has dug a deep hole in the street at the bottom of the hill. A Siberian tiger, escaped from the City Zoo, has taken up residence in the hole. You apply the brakes and lock your wheels at the top of the hill, where you are moving with a speed of 20 m/s. The inclined street in front of you is 40 m long.

A)Will you plunge into the hole and become the tiger's lunch, or do you skid to a stop before you reach the hole? (The coefficients of friction between your motorcycle tires and the wet pavement are μs=0.90 and μk=0.70.

Homework Equations

F_fr = μN
∑F = ma
∑F_net = ma = μN - mg sin70

The Attempt at a Solution

I combined all of the equations above canceling out the mass to get an acceleration of 2.452 m/s^2. I then used a kinematics equation V_x = V_o + at to find that the time to reach 0 m/s speed was 12.23 s. (This number seems large to me). I then used another kinematics equation Dx = v_ot + 1/2 at^2. I found the distance by the man to be something around 183m. This does not seem right to me, but I don't know where my logic is flawed. Could someone check me on this?

 

[collinsmark]

Homework Statement

The problem reads : You are riding your motorcycle one day down a wet street that slopes downward at an angle of 20 ∘ below the horizontal. As you start to ride down the hill, you notice a construction crew has dug a deep hole in the street at the bottom of the hill. A Siberian tiger, escaped from the City Zoo, has taken up residence in the hole. You apply the brakes and lock your wheels at the top of the hill, where you are moving with a speed of 20 m/s. The inclined street in front of you is 40 m long.

A)Will you plunge into the hole and become the tiger's lunch, or do you skid to a stop before you reach the hole? (The coefficients of friction between your motorcycle tires and the wet pavement are μs=0.90 and μk=0.70.

Homework Equations

F_fr = μN
∑F = ma
∑F_net = ma = μN - mg sin70

70? Isn't he angle 20^o with respect to the horizontal?

The Attempt at a Solution

I combined all of the equations above canceling out the mass to get an acceleration of 2.452 m/s^2.

You should show your work for how you got this number. Per the forum rules, we can't help you significantly if you don't show your work.

That said, I came up with a different value for the acceleration.

I then used a kinematics equation V_x = V_o + at to find that the time to reach 0 m/s speed was 12.23 s. (This number seems large to me). I then used another kinematics equation Dx = v_ot + 1/2 at^2. I found the distance by the man to be something around 183m. This does not seem right to me, but I don't know where my logic is flawed. Could someone check me on this?

While there's nothing technically wrong with solving for t first and then using that in a separate equation, you can save yourself some time by picking a different kinematics equation from the beginning.

There is a kinematics equation for uniform acceleration that is a function of initial and final velocities, acceleration and distance. Using that equation, you can find the rest of the answer in a single step.

It's not a huge issue in homework problems, but picking the right equation at the beginning can save you lots of precious time when taking tests.

 

[PhysicsDerp]

Since it is downward at an angle of 20 ∘ "below the horizontal" shouldn't my angle be 70 degrees when I draw my triangle? The angle at the bottom of the hill would be 20 I thought. Can you tell me if my net force equation is correct?

 

[collinsmark]

Since it is downward at an angle of 20 ∘ "below the horizontal" shouldn't my angle be 70 degrees when I draw my triangle? The angle at the bottom of the hill would be 20 I thought. Can you tell me if my net force equation is correct?

Yes, you can draw it like that.

But then the component of the gravitational force acting along the surface would involve the cosine(70) function instead of the sine function wouldn't it?