A 1130-kg car is held in place by a light cable on a very smooth (frictionless) ramp (Fig. E5.8). The cable makes an angle of 31.0° above the surface of the ramp, and the ramp itself rises at 25.0° above the horizontal. (b) Find the tension in the cable.

Video transcript

Hey, everybody, let's get going. So an 850 kg car is to be loaded on a car transporter. A 22 degree ramp measured above the horizontal is used for the process. A light cable making an angle of 28 degrees above the surface of the ramp is used to hold the car temporarily stationary on the ramp. Find the tension in the cable if the ramp is very smooth or frictionless. So we need to find our tension T in this cable. And we are going to draw a free body diagram to figure out What we have and how we can find what we need. So we have our mass 850 kg On a 22° incline measured from the horizontal. And then we have a cable, the tension force t that we need to find. And what else do we have? We have gravity MG, which we do know because we know G and we know the mass of the car and then we have a, an angle here as well. The angle that the cable makes with The top of the ramp is going to be degrees remove that okay. So, so we're giving two angles, we have a mass and we can also find MG by multiplying the mass with gravity. What other forces do we have? So let's split everything into Y and X forces. So we have a and we could do that by drawing a right triangle, of course. So we can draw our M G Y which is going to work at the same angle as our incline 22 degrees. And then we also have an MG X acting towards the left or in the negative direction. And then we also have our tension force which we can split into a Y component just going to be pulling against gravity and that'll be T Y. And then we also have an X component opposing gravity as well along this angle here. That's going to be our T X. And then we also have one more force whenever you have gravity pulling an object against another object, you of course have the reactive or normal force we'll call that in. So now we need to figure out how we can get T so because we have, we, we don't know any of our Y component values. We will apply Newton's second law in the X axis to determine to be, to be able to solve for the other forces as well. So the sum of forces and the extraction is of course equal to the acceleration times the mass in the Y direction. And now we can substitute here, we only have two forces. We have our positive tension force in the X axis and we have our negative gravitational force in the X axis and that will be equal to zero because we have attention going on. So now we can conclude that T of X is equal to MG X. Now, how do we find MG X? We know recall that MG X is just the result of multiplying our main force mg with the sign of our angle here 22 degrees, our angle up here because it's measured from the vertical 22 degrees. And then when you calculate that You should get 3100 20 newtons. And since that's our MG X value, we also know that T of X Is equal to 3120 Newtons. Now, so we know the X component of our tension and we can also apply a similar role from before to this. So T of X is equal to the total tension T times in this case, because Our angle is measured from the horizontal at 28°. It'll be the co sign of our angle. And now we can divide TFX by the co sign of 28. And that should give you a total tension force of 34 Newtons. So let's return to our answer choices and we can see that answer be 3534 newtons is the correct answer. Thank you.