The driver of a 2600 kg truck travelling at 13.1 m/s along a level but icy road steps on the brakes. The wheels lock and the tires slid along the icy surface for a distance of 26.2 m before coming to a stop. Calculate the coefficient of kinetic friction between the tires' surface and the ice.
7 years ago
Answered By Michael Z
So for this question the main thing to understand is that it is not the brakes that are doing the stopping but the kinetic friction since the wheels have locked up. To start this question we need to draw a force diagram. As you can see from the force diagram it will be only force of kinetic friction acting to slow the truck down. Once the truck has slowed down and stopped force of friction won't be acting on the truck anymore and it will be at a stand still.
1) we need to calculate the deceleration of this truck. luckily from our kinematics unit we learned this formula
Vf2= Vi2 + 2 a d ( rearrage to solve for a
2)apply newtons second law Fnet= ma to solve for the force of friction
3) Apply the friction formula FF= (u)FN ( note that FN=Fg=mg) and there you have it.
After the wheels have locked up, it is solely the frictional force between the ice and the wheels (which are now stationary) that is decelerating the truck. Let's draw a force diagram of the situation and identify all the important quantities. This I have done in the drawing below. Fr is the frictional force on the wheels of the the truck, which is initially (right after applying the brakes) moving to the right with a velocity, vi = 13.1m/s. The frictional force is acting in the opposite direction of its motion, until the truck comes to a rest. 'N' is the normal reaction force on the truck, 'm' is the mass, 'g' is gravitational acceleration.
The important assumption here is that the frictional force is CONSTANT during this time, from the moment the brakes are applied to the moment it comes to a stop. This is called kinetic friction, and it is related to the normal force via:
$\left|F_r\right|=\mu N$|Fr|=μN
Where $\mu$μ is the coefficient of friction that we are after (the bars around Fr means you only take the MAGNITUDE of the force, ignoring any minus sign). So in order to calculate this coefficient, we rearrange:
$\mu=\frac{\left|F_r\right|}{N}$μ=|Fr|N (1)
We already know what N is (it is mg, and we know what m is from the question). Thus we have to find Fr. To do this we first calculate acceleration, a. This we can do using the kinematic equation:
$v_f^2-v_i^2=2ad$vƒ2−vi2=2ad (2)
Where vf is the final velocity and vi is the initial velocity, and d is the distance travelled. The final velocity is zero (since the truck is at rest), and we are given distance and initial velocity in the question, so we just need to rearrange eqn.(2) and solve for a:
$a=\frac{-v_i^2}{2d}$a=−vi22d
Now all we have to do to get Fr is multiply the acceleration by mass (remember F =ma!), and plug the frictional force back into eqn.(1). There you have it!
7 years ago
Answered By Michael Z
So for this question the main thing to understand is that it is not the brakes that are doing the stopping but the kinetic friction since the wheels have locked up. To start this question we need to draw a force diagram. As you can see from the force diagram it will be only force of kinetic friction acting to slow the truck down. Once the truck has slowed down and stopped force of friction won't be acting on the truck anymore and it will be at a stand still.
1) we need to calculate the deceleration of this truck. luckily from our kinematics unit we learned this formula
Vf2= Vi2 + 2 a d ( rearrage to solve for a
2)apply newtons second law Fnet= ma to solve for the force of friction
3) Apply the friction formula FF= (u)FN ( note that FN=Fg=mg) and there you have it.
Attached Whiteboard:
Play Drawing7 years ago
Answered By Kazi A
After the wheels have locked up, it is solely the frictional force between the ice and the wheels (which are now stationary) that is decelerating the truck. Let's draw a force diagram of the situation and identify all the important quantities. This I have done in the drawing below. Fr is the frictional force on the wheels of the the truck, which is initially (right after applying the brakes) moving to the right with a velocity, vi = 13.1m/s. The frictional force is acting in the opposite direction of its motion, until the truck comes to a rest. 'N' is the normal reaction force on the truck, 'm' is the mass, 'g' is gravitational acceleration.
The important assumption here is that the frictional force is CONSTANT during this time, from the moment the brakes are applied to the moment it comes to a stop. This is called kinetic friction, and it is related to the normal force via:
$\left|F_r\right|=\mu N$|Fr|=μN
Where $\mu$μ is the coefficient of friction that we are after (the bars around Fr means you only take the MAGNITUDE of the force, ignoring any minus sign). So in order to calculate this coefficient, we rearrange:
$\mu=\frac{\left|F_r\right|}{N}$μ=|Fr|N (1)
We already know what N is (it is mg, and we know what m is from the question). Thus we have to find Fr. To do this we first calculate acceleration, a. This we can do using the kinematic equation:
$v_f^2-v_i^2=2ad$vƒ 2−vi2=2ad (2)
Where vf is the final velocity and vi is the initial velocity, and d is the distance travelled. The final velocity is zero (since the truck is at rest), and we are given distance and initial velocity in the question, so we just need to rearrange eqn.(2) and solve for a:
$a=\frac{-v_i^2}{2d}$a=−vi22d
Now all we have to do to get Fr is multiply the acceleration by mass (remember F =ma!), and plug the frictional force back into eqn.(1). There you have it!
Attached Whiteboard:
Play Drawing