A uniform ladder is 10m long and weighs 400 N. It rests with its upper end again

Question

A uniform ladder is 10m long and weighs 400 N. It rests with its upper end against a frictionless

vertical wall. Its lower end rests on the ground and is prevented from slipping by a peg driven

into the ground. The ladder makes a 30? angle with the horizontal. The magnitude of the

force exerted on the peg by the ladder is:

A. zero

B. 200N

C. 400N

D. 470N

E. 670N

ans: D

A uniform ladder is 10m long and weighs 400 N. It rests with its upper end against a frictionless vertical wall. Its lower end rests on the ground and is prevented from slipping by a peg driven into the ground. The ladder makes a 30? angle with the horizontal. The magnitude of the force exerted on the peg by the ladder is: zero 200N 400N 470N 670N

Explanation / Answer

First, you want to draw your forces on the diagram. The weight of the ladder acts in the middle of the ladder (center of mass), and since the wall is frictionless, it only exerts a normal force that points left.

The peg exerts a force both right (to cancel the wall force) and a force up (to cancel the weight of the ladder).

Now we have our equilibrium equations:

Sum of forces in the x direction = 0
Sum of forces in the y direction = 0
Sum of torques = 0

and we have three unknown forces. Now, let's take a look at the forces in the y-direction. The force of the peg upward must be equal to the weight of the ladder (since the other two forces are in the x-direction). So we know that the force of the peg pushing the ladder upwards is 400 N.

Next we can look at sum of torques. Here you get to pick where you want your origin to be. I would suggest making it at the peg; that way you only have two torques to deal with. We don't know the length of the ladder, so let's just call it 2L.

In the clockwise direction, the torque is L * 400cos(30) [remember, you only look at the component of the force that is perpendicular to the distance from the origin to the force]

In the counterclockwise direction, the torque is 2L*F*sin(30). When you set these equal to each other, the L's cancel out and you can solve for F, which is 200*sqrt(3). This is the force the wall exerts on the ladder.

Last, you can look the sum of forces in the x-direction. Since there are only two, the force of the peg in the x-direction has to be equal (in magnitude) to the force of the wall on the ladder.

Now that you have the two forces on the peg (400 and 200sqrt(3) ), the last step is to figure out the magnitude of the overall force.

magnitude = sqrt(Fx^2 + Fy^2)

When we plug in our numbers, we get:

F = sqrt[ (200sqrt(3))^2 + 400^2]
F = sqrt[ 280000 ]

F = 200sqrt(7) Newtons

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