A 4000 kg truck is parked on a 15 degrees slope. How big is the friction force on the truck? The coefficient of static friction between the tires and the road is 0.90.

Video transcript

Mhm Hey, everyone in this problem, a 1000 kg block is resting on a plane inclined at degrees. We're asked to calculate the force of friction experienced by the block. If us between the block and the surface of the slope is 0.87, we're given four answer choices. Option A 3240 noons. Option B 3532 Newtons, option C 3420 Newton and option D Newtons. Now let's go ahead and draw this out. So what do we have? We have an incline and this makes 20 degree angle with the horizontal and we have our block that is resting on that plane and our block has a mass of 1000 kg. Now let's try a pre body diagram of our block acting upwards perpendicular to the plane is the normal force acting straight down. OK. Vertically is the force of gravity and acting to the left parallel to the plane. The inclined plane is the force of static friction. Now, why have I called this static friction and not kinetic friction? Well, we're told that the block is at rest. OK. If it's at rest, we're dealing with static friction. Now, it's pointing to the left because friction always opposes the motion. So if we were to let this block go with a large enough force of gravity with a small enough friction, it would slide down the incline. Ok. So the friction opposes that motion and points up the incline. Now, we wanna take a tilted axis here. So we want the axis when we're talking about our free body diagram to be parallel and perpendicular to our inclined plane. So what that means is we need to break this force of gravity into its X and Y components according to this tilted axis. If we do that, we get the X component of the force of gravity F G X pointing to the right. OK. In terms of our inclined plane and we have the force of gravity, Y component pointing down in our plane and the angle between the force of gravity and the X component is going to be 20 degrees and it's not my mistake. It is not going to be 20 degrees. OK. The angle between F G and the Y component is going to be degrees. OK. It can be easy to mix that up when you're working with a tilted axis. So always double check. OK. So we have the angle between the force of gravity and the Y component is 20 degrees. All right. Great. Now we're gonna take up into the right in our tilted axis as positive. OK. So that means perpendicular, parallel up and to the right now this block is at rest. What that tells us is that the, some of the forces is equal to zero. Right now, we're interested in friction. So we're gonna look at the sum of the forces in the X direction. What forces do we have in the direction? Well, we have the force of gravity acting in the X direction that's in the positive direction. And we have the force of static friction acting in the negative X friction. So we have F G X minus F S is equal to zero that tells us that the force of static friction we're looking for is going to be equal to the X component of the force of gravity. Now, this X component of F G is on the opposite side. So we're gonna relate it through the sign of the angle. So we get signed of 20 degrees multiplied by the mass multiplied by G, the gravitational acceleration. OK? Because that is F G substituting in our values sine of 20 degrees multiplied by 1000 kg multiplied by 9.8 m per second squared. And we get the F S, the force of static friction is equal to 0.8 in the unit. Here we have kilogram meter per second squared. That is equivalent to a Newton. And so we get the unit of noon, which is what we would expect for a force. And comparing this to our answer choices, we see that they've rounded to the nearest Newton. And so the force of friction that we found, which is the force of static friction because the block is at rest is approximately 3352 newtons, which corresponds with answer choice. D Thanks everyone for watching. I hope this video helped see you in the next one.