Consider a large truck carrying a heavy load, such as steel...

Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slide forward, crushing the cab, if the truck stops suddenly in an accident or even in braking. Assume, for example, a 15,000-kg load sits on the flatbed of a 20,000-kg truck moving at 10.0 m/s. Assume the load is not tied down to the truck and has a coefficient of static friction of 0.550 with the truck bed. (a) Calculate the minimum stopping distance for which the load will not slide forward relative to the truck. m (b) Is any piece of data unnecessary for the solution? (Select all that apply.) mass of the load mass of the truck velocity coefficient of static friction

Transcript

00:01 So in this problem, we have some truck of 20 ,000 kilograms with a load on it of 13 ,000 kilograms.

00:08 This truck is going 10 meters per second, but decelerating and stopping.

00:18 What we want to find is with the maximum acceleration so that we can get the lowest stopping distance so that this load of, say, steel is not going to hit the driver's cap.

00:31 So the reason why this load would stay connected to the truck is because of friction.

00:38 There is some friction force on the box of load here.

00:45 We know that friction is equal to the coefficient of static friction times the normal force.

00:51 Static coefficient friction because we want this object to remain stationary relative to the thing is on, which is the truck bed.

01:00 And we know that mu here is equal to 0 .4.

01:08 So what we want to do is find the maximum acceleration so that this object doesn't move.

01:16 And so to do that, we'll draw a free by diagram of the steel load.

01:25 We have the force of friction on it.

01:33 And then we also have the weight, which is equal to mass times gravity from newton's second law.

01:42 And as a reaction to the weight, there is a normal force from the truck bed, which we put like such, and then we can draw our typical x and y axis.

01:59 So what we're really interested in is finding friction, and to do that, we don't know n, but we can find it from here.

02:07 Using newton's second law in the y direction here, we say that some forces of y is equal to zero, so that the box stays stationary relative to the truck bed.

02:18 And then we can directly write n minus m g is equal to zero n is positive in the positive y m g is negative because it's in the negative y and so you see that n is equal to m g and we can call it say m of s for steel so in the x direction we have acceleration so instead of equal to zero the sum of forces in the x direction is equal to the massive steel times its acceleration because it's stationary, we know that the acceleration of the truck and the box will be the same.

02:58 So the only force we have in the x direction is the friction.

03:03 So ma is equal to negative mu -s -n, because the friction is in the negative direction by this diagram.

03:12 We can plug in what we know for n, mu -s, and then times the mass times gravity.

03:24 We see that the mass cancels on each side of this equation.

03:27 So we are left with the acceleration is equal to negative m -r -s times g, which we can then plug in the values for.

03:41 We know the coefficient of friction is 0 .4, and gravity is negative 9 .81 meters per second squared due to the acceleration due to gravity on earth.

03:56 And so that works out to an acceleration of 3 .924.

04:05 Meters per second squared.

04:10 Although actually what we should do here specifically is we've taken care of the sign of g by the arrow so we can leave that as positive, which means his answer for acceleration is negative, which makes sense because by our axis system, exclamation is to the negative direction.

04:32 So now we have the acceleration...

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