Understanding Why Tires Grip the Road Better on Level Ground

[Dooga Blackrazor]

1. Why do tires grip the road better on level ground than they do going uphill or downhill?

I have some ideas, but I'm not sure how to word them, or if they are correct. Any suggestions would be appreciated.

2. A mass M(1) = 3.5 kg, rests on a horizontal table and is attached by ropes to masses M(2) = 1.5 kg and M(3) = 2.5 kg. After is is released, what are the magnitude and direction of the acceleration of M(1)? Ignore friction.

Answer: I calculated 1.14 m/s^2 to the right. The answer is known to be between 1-2. If anyone can confirm my calculation it would be appreciated.

~Thanks.

 

[Andrew Mason]

1. Why do tires grip the road better on level ground than they do going uphill or downhill?

I have some ideas, but I'm not sure how to word them, or if they are correct. Any suggestions would be appreciated.

The 'grip' depends on the normal force which is opposite to the component of gravity perpendicular to the road. When the vehicle is inclined by an angle $\theta$ from the horizontal, the normal force changes by a factor of $cos\theta$.

2. A mass M(1) = 3.5 kg, rests on a horizontal table and is attached by ropes to masses M(2) = 1.5 kg and M(3) = 2.5 kg. After is is released, what are the magnitude and direction of the acceleration of M(1)? Ignore friction.

I gather that M(1) hangs down off the table to the left and M(3) hangs down on the right and there is some kind of pulley with each (you should provide all the details).

Answer: I calculated 1.14 m/s^2 to the right. The answer is known to be between 1-2. If anyone can confirm my calculation it would be appreciated.

I get a different answer. Why don't you show what you did.

AM

 

[Dooga Blackrazor]

Thanks:

For question two I did the following:

Object 1:

Fnety = 0
Fnetx = T = M(1)a

Object 2:

T- W(2) = -M(2)a
T = W(2) - M(2)a
T = M(1)a
M(1)a = W(2) - M(2)a

a = M(2)g % M(1) + M(2)

Then I substituted in #'s and repeated the procedure for M(3). I took the result from the left acceleration and subtracted it from the larger right acceleration.

 

[Andrew Mason]

Object 1:

Fnety = 0
Fnetx = T = M(1)a

Object 2:

T- W(2) = -M(2)a
T = W(2) - M(2)a

What is W(2)? Weight? Why is it important?

T = M(1)a
M(1)a = W(2) - M(2)a

a = M(2)g % M(1) + M(2)

What is %? This makes no sense at all to me.

Then I substituted in #'s and repeated the procedure for M(3). I took the result from the left acceleration and subtracted it from the larger right acceleration.

Since the accelerations of all three masses are equal just treat all three masses as one mass. Then use the fact that the net force is (M3 - M1)g. It is that simple.

AM

 

[Dooga Blackrazor]

W(2) = The weight of object two. The Fnet of object two is determined by -M(2)a because it is going down. The % is being used as division.

How can you make the masses equal when M(1) is ona table and the other masses are hanging from it. The higher weight of mass 3 is what pulls M(1) to the right, but M(2) has an acceleration which slows the acceleration to the right.

 

[Andrew Mason]

W(2) = The weight of object two. The Fnet of object two is determined by -M(2)a because it is going down. The % is being used as division.

How can you make the masses equal when M(1) is ona table and the other masses are hanging from it. The higher weight of mass 3 is what pulls M(1) to the right, but M(2) has an acceleration which slows the acceleration to the right.

It is the difference in the weight of M3 and M1 that provides the acceleration to the system.

$$(M_1 + M_2 + M_3)a = M_3g - M_1g$$

$$a = g\frac{M_3 - M_1}{M_1 + M_2 + M_3} = 9.8 * \frac{1}{7.5} = 1.31 m/s^2$$

AM

 

[Dooga Blackrazor]

Ok, I've got it now. Thanks!