# Calculating Minimum Stopping Distance for Truck & Box

• NAkid
In summary: I've tried a bunch of things so far, but I just can't seem to get it.In summary, the truck can stop at the least distance of 104.2 meters and ensure that the box does not slide.
NAkid

## Homework Statement

The coefficent of static friction between the floor of a truck and a box resting on it is 0.39. The truck is traveling at 87.8 km/hr. What is the least distance in which the truck can stop and ensure that the box does not slide?

Fs=Us*N

## The Attempt at a Solution

I really am not sure how to work this out. I attempted using formula Vf^=V0^2 + 2adx but I don't think that's right. Help please!

Sorry, my other question is this:
M1 and M2 are two masses connected as shown. The pulley is light and frictionless. Find the mass M1, given that M2 (7.00kg) accelerates downwards at 3.43m/s2, that theta is 20.0o, and that muk is 0.390.

I'm pretty sure I set it up correctly:
for block 2 i said, T-mg=ma --> T=ma+mg
for block 1, T-fk-mgsin(theta)=ma AND N-mgcos(theta)=0 (no acceleration in vertical direction)

also, i know that m2>m1 for m2 to accelerate downwards

basically i plugged everything in but the answer is still wrong..

Last edited:
NAkid said:

## Homework Statement

The coefficent of static friction between the floor of a truck and a box resting on it is 0.39. The truck is traveling at 87.8 km/hr. What is the least distance in which the truck can stop and ensure that the box does not slide?

Fs=Us*N

## The Attempt at a Solution

I really am not sure how to work this out. I attempted using formula Vf^=V0^2 + 2adx but I don't think that's right. Help please!

I'm also having trouble with this problem. Here is what I tried (which didn't work):
First my numbers are different. Us=0.25 and v=81.4 km/hr.

First I converted 81.4 km/hr to 22.61 m/s.

Fs = Us*N => ma = Us*mg => a = Us*g => a = (0.25)(-9.81) = -2.4525 m/s^2
v^2 = v0^2 + 2a(delta x) => 0 = 22.61 + 2(-2.4525)(delta x) => delta x = 4.61 m

This is the wrong answer according to LONCAPA, so I don't really know what to do now.

anastasiaw said:
First I converted 81.4 km/hr to 22.61 m/s.

Fs = Us*N => ma = Us*mg => a = Us*g => a = (0.25)(-9.81) = -2.4525 m/s^2
v^2 = v0^2 + 2a(delta x) => 0 = 22.61 + 2(-2.4525)(delta x) => delta x = 4.61 m

This is the wrong answer according to LONCAPA, so I don't really know what to do now.

You forgot to square the initial velocity in the second equation.

Last edited:
spizma said:
You forgot to square the initial velocity in the second equation.

Oh simple mistakes... they kill me.

I have the right answer now... 104.2 m

NAkid said:
Sorry, my other question is this:
M1 and M2 are two masses connected as shown. The pulley is light and frictionless. Find the mass M1, given that M2 (7.00kg) accelerates downwards at 3.43m/s2, that theta is 20.0o, and that muk is 0.390.

I'm pretty sure I set it up correctly:
for block 2 i said, T-mg=ma --> T=ma+mg
for block 1, T-fk-mgsin(theta)=ma AND N-mgcos(theta)=0 (no acceleration in vertical direction)

also, i know that m2>m1 for m2 to accelerate downwards

basically i plugged everything in but the answer is still wrong..

I'm also having trouble with this one if anyone has any advice to offer.

## 1. What is the formula for calculating minimum stopping distance for a truck and box?

The formula for calculating minimum stopping distance for a truck and box is d = (V^2)/(2μg) + Vt, where d is the stopping distance, V is the initial velocity, μ is the coefficient of friction, g is the acceleration due to gravity, and t is the reaction time of the driver.

## 2. How do you determine the coefficient of friction for a truck and box?

The coefficient of friction for a truck and box can be determined by conducting a skid test, where the truck is driven at a certain speed and then the brakes are applied until the truck comes to a stop. The distance it travels during braking is measured and used in the formula to calculate the coefficient of friction.

## 3. What factors can affect the minimum stopping distance for a truck and box?

There are several factors that can affect the minimum stopping distance for a truck and box. These include the initial velocity of the truck, the weight and load of the truck and box, the condition of the brakes and tires, the road surface and weather conditions, and the reaction time of the driver.

## 4. Can the minimum stopping distance for a truck and box be reduced?

Yes, the minimum stopping distance for a truck and box can be reduced by maintaining the truck and its components in good condition, such as regular brake and tire checks, and by following safe driving practices. It is also important for the driver to have a shorter reaction time, which can be achieved through proper training and alertness.

## 5. How accurate is the minimum stopping distance calculated using the formula?

The minimum stopping distance calculated using the formula is an estimate and may not be completely accurate in real-life situations. Factors such as driver error, unexpected obstacles, and varying road conditions can affect the actual stopping distance. It is important for drivers to always be cautious and leave enough space for braking to ensure safety on the road.

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