Find the force required to keep a 75-lb sled from sliding down an incline that makes an angle of 27° with the horizontal. (Assume there is no friction.)

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Hey, everyone here we are given a cubicle box is kept on an incline surface that makes an angle of 35 degrees with the horizontal, determine the force required to keep the box from sliding down. If the weight of the box is 525 kg, we need to assume that there is no friction. We have four answer choice options, answer a 430 kg. B 525 kg, see 103 kg and D 301 kg. To begin this problem, we need to create a sketch containing our given information. So we can start by simply drawing our horizontal line. And from there, we can draw our inclined surface by just extending a line upward from our horizontal. And we are given that this inclined surface has an angle of 35 degrees with the horizontal. So we can go ahead and label that angle. Now, we know we have a box on this inclined surface so we can go ahead and include that on our sketch and we have an unknown force f keeping our box in place. So we can show this force as an arrow going from our box upward along the incline surface. And we can label this unknown force as force F. And lastly, we are given that the box weighs 525 kg. So we represent this weight force as another arrow which extends from the box directly downward. So that this arrow is perpendicular to the horizontal. And we just need to label this arrow as 525 kg. Now, to find our unknown force F, we need to utilize triangles with our sketch. So first, we can extend a line downward from the end of our inclined surface directly down to the horizontal. And we know that this will form a 90 degree angle with the horizontal, which means we have completed a large right triangle. However, this triangle alone doesn't give us enough information to find f. So we need to recreate this triangle utilizing our weight force. So now we extend another line from our box downward except this line must be perpendicular to the inclined surface. And now, since this line is perpendicular to the surface and our weight force is perpendicular to the horizontal, this tells us that the angle between these two lines is our given angle 35 degrees. And now we just want to finish our triangle by utilizing our unknown force. So we see that this force forms a perpendicular line with the line we have just drawn in since the force is parallel to the surface. And so the last side of our triangle will be this force. So we extend an arrow from our weight force to our new line and this forms our 90 degree angle. And we just want to label this side as force F. And now looking at our new triangle, we notice we have one angle known and we also know the value for the hypotenuse. So we have enough information to find our force F. So first, we want to label the top point of our triangle as point A, then the left point will be point B and the right point will be point C. And now with our triangle ABC, we can see that sine of angle A is equal to sin length BC divided by length B. Now just substituting in our known information, we have sine of 35 degrees is equal to F divided by 525. Now solving for F, we have that F is equal to 525 multiplied by s of 35 degrees. Now, just evaluating, we find that F is approximately equal to 301.1276 kg. And now just rounding to the nearest whole number. Here, we find that our unknown force F is approximately equal to 301 kg. And with our final rounded answer here we are left with answer choice D where again, we found our Unknown forest is 301 kg. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

Find the force required to keep a 75-lb sled from sliding down an incline that makes an angle of 27° with the horizontal. (Assume there is no friction.)

Verified Solution

Key Concepts

Inclined Plane

Components of Forces

Newton's Second Law