<p>Can someone please help me on this question? Thanks!</p>
<p>A 5.0 m long ladder leans against a wall, touching it at a point 3.8 m above the ground. The ladder is uniform with a mass of 15kg. The wall is frictionless. What is the horizontal normal force acting on the ladder? (meaning the normal force by the wall)</p>
<p>I know you use torque…but I just can’t seem to get the right equation…</p>
<p>Oh…and if you could explain your line of thinking, I would be really grateful.</p>
<p>There are 4 forces here:</p>
<p>Normal force n acting horizontally where the top of the ladder contacts the wall.</p>
<p>Friction force where the ladder contacts the ground, equal but opposite to the normal force.</p>
<p>The downward weight w of the ladder, at one half the ladder’s length (in other words, 2.5 m).</p>
<p>Normal force directed upward, where the ladder touches the ground.</p>
<p>I’m going to use the angle ‘theta’ as the angle the ladder makes with the ground (you can figure out what theta is because you know that the ladder/hypotenuse is 5.0 and the height is 3.8).</p>
<p>We want net torque to be zero, and we set the base of the ladder (where it touches the ground) to be 0, and the top of the ladder (where it touches the wall) to be 5. Since the frictional force and vertical normal force are both at 0, we can leave them out of our calculations.</p>
<p>Then we’re just left to figure out the components of the weight and the horizontal normal force at the top that are perpendicular to the axis of rotation (aka, the ladder).</p>
<p>Net Torque=0= (2.5m)(-w)cos(theta) + (5.0m)(n)sin(theta)</p>
<p>where you’re solving for “n”.</p>
<p>ahh chapter 9 giancoli</p>
<p>Thanks GoldShadow!</p>
<p>@thecalccobra: Actually, we’re not using the giancoli book…We used it last year as a supplement text for phys h. XDD I like that book a whole lot better than any of the “official” books for our school.</p>
<p>I think thats chapter 8* giancoli</p>