A uniform ladder of length l and mass m leans against a frictionless vertical wall, making an angle of 53° with the horizontal. the coefficient of static friction between the ladder and the ground is 0.32. if your mass is four times that of the ladder, how high can you climb before the ladder begins to slip?

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09-11-2017
Physics

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A uniform ladder of length l and mass m leans against a frictionless vertical wall, making an angle of 53° with the horizontal. the coefficient of static friction between the ladder and the ground is 0.32. if your mass is four times that of the ladder, how high can you climb before the ladder begins to slip?

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W0lf93 W0lf93 · Answer 1 · 2017-11-21T00:51:13+01:00

Answer: Normal forces: N1 = normal force acting horizontal on top end of ladder touching the frictionless wall N2 = normal force acting vertically up on the bottom of the ladder touching the ground with friction Friction forces: Ff = friction force acting horizontal towards the vertical wall on the bottom of the ladder. Ff will be equal to mu (coefficient of static friction) * N2 (Ff = mu*N2) Weight Forces: W = weight of the ladder acting at the centroid of the ladder at a distance L/2 from the base of the ladder along the line of the ladder WP = 4*W = weight of the person acting at an arbitrary distance x1 from the base of the ladder along the line of the ladder If you can get this picture the rest should be easy! Now to solve for some of our forces in terms of our variables. By summing the forces in the vertical or often the y-direction we can obtain N2. The sum will equal zero because the system is static (no acceleration is occurring). 0 = N2 - W - WP W and WP are negative because they act down, while N2 acts up and is positive for my connotation. Therefore N2 = W + WP and WP = 4*W therefore N2 = 5*W As stated above we know that Ff = mu*N2 therefore Ff = mu*5*W with a simple substitution. Now to complete the problem we will need to sum moments about a point. There are several ways of doing this, I chose to pick a point that cleverly excuses N1 and N2 from our moment equation. I did so by choosing the point that N1 and N2 act through above the ladder. Its difficult to expresses my positive/negative connotations for this section but forces that create a moment around this point CCW will be positive and those that create a CC moment about this point will be negative. Again the sum of the moments will add up to zero as the system is static. Before I sum moments let me define some lengths that I will need. d1 = the vertical length from the ground to where the ladder touches the vertical wall. With simple trig. we can see that d1 = L*sin(beta) d2 = the horizontal distance from the base of the ladder to the line of action of the person's weight. Again with simple trig. we can determine that d2 = x1*cos(beta) d3 = the horizontal distance from the base of the ladder to the line of action of the ladder's weight. Same logic shows that d3 = L/2*cos(beta) Now to sum the moments about my point described above. 0 = 4*W*d3 + W*d3 - Ff*d1 The first term is the moment created by the person, the second term is the moment created by the ladder, and the third term is the moment created by the friction. Substituting our known values into the terms above we get: 0 = 4*W*x1*cos(beta) + W *L/2*cos(beta) - mu*5*W*L*sin(beta) First off we can see each term contains a W; and W does not equal 0 therefore we can divide each term by W and find that it will no longer be contained in the equation: 0 = 4*x1*cos(beta) + L/2*cos(beta) - mu*5*L*sin(beta) Now I will begin solving the equation for x1 Step one: move all terms to one side that do not contain an x1 mu*5*L*sin(beta) - L/2 *cos(beta) = 4*x1*cos(beta) Step two: divide by 4*cos(beta) to get x1 by itself x1 = mu*5/4*L*tan(beta) - L/8 Step three (probably unnecessary): plug in given values mu = 0.4 and beta = 53 degrees x1 = 0.66*L - L/8 x1 = 0.54*L or the person can climb slightly passed L/2 on the line of the ladder without the ladder slipping.