A 1500 kg car takes a 50-m-radius unbanked curve at 15 m/s. What is the size of the friction force on the car?

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Hey, everyone in this problem, a truck with a mass of 2600 kg is driving on a horizontal unmatched curve at the radius of 38 m at a constant speed of 18 m per second. We are asked to find the magnitude of the force of friction acting on the truck that is required to keep it moving in a circular path. We're given four answer choices. Option A 4.28 times 10 to the exponent three newtons. Option B 6.36 times 10 to the exponent six newtons. Option C 2.22 times 10 to the exponent four newtons and option D 3.42 times 10 to the exponent four newtons. So we're asked to find a force of friction. Let's go ahead and draw a free body diagram of all of the forces acting so that we can get a sense of how to calculate this force we're looking for. So we have our truck, it is on the road. So we know that there's gonna be a normal force pointing upwards. OK. This is a flat curve. So that's pointing straight vertically. We have the force of gravity acting downwards in the horizontal direction, we have the force of static friction. Now, we're gonna put that force pointing to the right and that's gonna be F S. Now, if the force of static friction is pointing to the right, that means it's stopping the car or the truck, preventing it from sliding out to the left. OK. And so the acceleration, the centripetal acceleration is actually gonna point to the right as well. OK. That force of friction pointing the right is keeping that truck moving along that circular path with the acceleration pointing inwards. I recall from Newton's second loss. We have that the sum of the forces and this is the centripetal forces is equal to the mass multiplied by the centripetal acceleration ac. Now, in that horizontal direction, we have only the force of static friction we're gonna take to the right to be positive. And so that force of static friction is gonna be positive F S and this is equal to the mass multiplied by this centripetal acceleration. Now, we aren't given any information about the acceleration of this truck, but we are given information about its speed. Now, let's recall that we can write this in triple acceleration ac as V squared divided by R. So now we have this force of friction we're trying to calculate is equal to M V squared divided by R. We know the mass, we know the speed V and we know the radius of this flat curve. So now it's just a matter of substituting in these values and working out the result, the mass is 2600 kg. That is multiplied by V 18 m per second, all squared divided by the radius R which is 38 m. So our force of static friction is equal to 2600 kg multiplied by 18 m per second squared. Divided by simplifying the numerator. We get 800 and 42, 400 kg meter squared per second squared. And that is divided by 38 m. Mhm In the numerator, we have meters squared. We're dividing by meter. So the meter in the denominator divides out and one of the meters in the numerator divides out. We're left with kilogram meters per second squared which recall is equivalent to a nude. When we work this out, we get that this is equal to 22, 0.42 newtons. Our answer choices are written in scientific notation with three significant digits. We can write this as 2.22 times 10 to the exponent four. OK. We have to move that decimal place to the left to get 2.22. OK. So 2.22 times 10 to the exponent four noons. And that is that force of friction that is keeping the truck moving in a circular path. If we compare this to our answer choices. We see that this corresponds with answer choice C that's it for this one. Thanks everyone for watching. See you in the next video.

A 1500 kg car takes a 50-m-radius unbanked curve at 15 m/s. What is the size of the friction force on the car?

Verified Solution

Key Concepts

Centripetal Force

Friction Force

Newton's Second Law of Motion