Two children pull a third child on a snow saucer sled exerting forces 𝐅1 and 𝐅2 as shown from ab

Two children pull a third child on a snow saucer sled...

Key Concepts

Newton's Second Law

This principle states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F_net = m*a). It is the fundamental relationship used to determine how forces affect the motion of an object, making it essential for solving problems involving the acceleration of a mass when forces are applied.

Vector Addition

When multiple forces are acting on an object from different directions, they must be combined using vector addition. This process involves decomposing each force into its components and summing these components along each coordinate axis to determine the total or net force acting on the object.

Free-Body Diagrams

A free-body diagram is a graphical representation that isolates an object and shows all the forces acting on it. Creating such diagrams helps to visualize the forces, correctly assign their directions, and set up the equations necessary for solving problems involving forces and motion.

Friction

Friction is a force that opposes the relative motion between contacting surfaces. In problems where friction is present but its magnitude or direction is not explicitly given, it is understood to act opposite to the net applied force, affecting the net force available to accelerate the object.

Transcript

00:02 For this problem, we need to solve for the acceleration of the sled child system.

00:10 So there's a child riding a sled.

00:14 So this sled child system is being pulled by two other persons and is experiencing a friction force from the ground.

00:23 So to better visualize the problem, so let's draw a sketch for the problem.

00:29 So suppose this is the sled.

00:34 With the child so this sled is being pulled by two persons so we have your two forces so one is in this direction and then and it's 45 degrees from the horizontal and then the other one is in this direction and is 30 degrees from the horizontal and as said earlier, the sled is experiencing a friction force.

01:22 So we actually don't know the direction of the friction force.

01:27 It's not specified.

01:29 So according to the problem, it will depend on the direction of the total applied force.

01:34 It's just opposite the direction of the total applied force.

01:40 So sorry we have here f1 applied let's label this f applied one and then f applied two so we have the values of this forces so f applied one is 10 neutons and then f applied two is eight newtons and then the friction force is also given so it's 7 .5 neutons so the mass of this sled child system is also given.

02:20 So the mass is actually 49 kilograms.

02:29 So again, we need to determine the acceleration.

02:33 So the acceleration of the sled child system.

02:38 So for the free body diagram, you can just simply copy our sketch.

02:47 So we have f applied force here so let's also indicate the angle so this is 45 degrees and then the other one f applied to uh 30 degrees then tentative direction of friction so here so this is our free body diagram so since the two applied forces are exerted at a certain angle, so they're not totally or they're not along the horizontal or along the vertical, so we need to resolve them to their components.

03:30 So let's draw that in our free body diagram.

03:33 So f applied one here will have a vertical component.

03:40 Let's write f applied one y.

03:43 Then also a horizontal component so f applied one x and then of course f applied two also have its own components so it will have a vertical component like this so f applied to sub y and then horizontal component f applied to sub x okay so that completes our free body diagram so to determine the or determine the equation for the x and y components will be using our knowledge of trigonometry so let's start first with f applied one so notice that we actually formed here a right triangle right here.

04:55 So based on the position of our angle, so here, if applied 1x is the adjacent angle, and of course f applied sub 1 is the hypotenuse.

05:15 So hence, we can use the definition for cosine, which is cosine, equals adjacent side over the hypotenuse.

05:31 So, in our case, cosine data equals our adjacent side, as said earlier, is the x component.

05:40 So f applied 1 subx over f applied sub 1.

05:48 Hence, f applied sub 1, x is equal to f applied sub 1.

05:58 1 times the times the cosine of the given angle so that's how we'll get the x component of the f applied 1 now for the for f applied sub 1 y or for the y component of f applied sub 1 so again looking at our angle so the y component or f applied applied sub 1 y is actually the opposite side.

06:38 So it's the opposite side.

06:40 Hence, we can use the definition of sign, which is sine theta equals opposite over the hypotenuse.

06:50 So in our case, that's sine theta equals the y component f applied sub 1 y over f applied sub 1.

06:59 So hence, our y component is equal to f applied sub 1 y equals f applied sub 1 times the sign of the given angle.

07:19 Okay, so that's how we get the y component.

07:23 So this goes also for the components of f sub 2.

07:27 If you look at the right triangle form, so this is all.

07:33 Also applicable for f applied sub 2.

07:40 Okay, so now that we've determined the equation for the components, we can now proceed with calculating for the net applied force or for the total applied force.

07:54 So along x, so along x, sorry, so the net force along x, summation f sub x equals so going back to our figure so we have two four two forces so this regarding first the friction so both x components of the two applied forces are directed to the right so hence our summation would be so f applied sub one xx plus f applied sub 2 x and we know that this are equal to the x the x component is equal to f applied sub 1 cosine of 45 so since for the applied force 1 it makes 45 degrees with the horizontal so cosine 45 degrees plus f applied sub 2 cosine 30 degrees so substituting the values so we have here so f applied 1 is 10 x .0 .000.

09:39 45 degrees plus f applied to is 8 newtons cosine 30 degrees.

09:46 So input this in your calculator...

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