A uniform ladder with mass m2�2 and length L� rests against a smooth wall. (Figure 1)A do-it-yourself enthusiast of mass m1�1 stands on the ladder a distance d� from the bottom (measured along the ladder). The ladder makes an angle θ� with the ground. There is no friction between the wall and the ladder, but there is a frictional force of magnitude f� between the floor and the ladder. N1�1 is the magnitude of the normal force exerted by the wall on the ladder, and N2�2 is the magnitude of the normal force exerted by the ground on the ladder. Throughout the problem, consider counterclockwise torques to be positive. None of your answers should involve π� (i.e., simplify your trig functions). Figure 1 of 1 Part A Part complete Part B Suppose that the actual coefficent of friction is one and a half times as large as the value of μmin�min. That is, μs=(3/2)μmin�s=(3/2)�min. Under these circumstances, what is the magnitude of the force of friction f� that the floor applies to the ladder? Express your answer in terms of m1�1, m2�2, d�, L�, g�, and θ�. Remember to pay attention to the relation of force and μs�s. View Available Hint(s)for Part B Hint 1for Part B. Relation between frictional force and μs�s The force of friction, in this case, is not equal to μsN2�s�2. In general, Ffriction<μsFnormal�friction<�s�normal. Only when the object is on the verge of slipping does Ffriction=μsFnormal�friction=�s�normal. To find f� recall that the torques about the point at which the ladder touches the wall must be zero. Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value type f� =
A uniform ladder with mass m2�2 and length L� rests against a smooth wall. (Figure 1)A do-it-yourself enthusiast of mass m1�1 stands on the ladder a distance d� from the bottom (measured along the ladder). The ladder makes an angle θ� with the ground. There is no friction between the wall and the ladder, but there is a frictional force of magnitude f� between the floor and the ladder. N1�1 is the magnitude of the normal force exerted by the wall on the ladder, and N2�2 is the magnitude of the normal force exerted by the ground on the ladder. Throughout the problem, consider counterclockwise torques to be positive. None of your answers should involve π� (i.e., simplify your trig functions). Figure 1 of 1 Part A Part complete Part B Suppose that the actual coefficent of friction is one and a half times as large as the value of μmin�min. That is, μs=(3/2)μmin�s=(3/2)�min. Under these circumstances, what is the magnitude of the force of friction f� that the floor applies to the ladder? Express your answer in terms of m1�1, m2�2, d�, L�, g�, and θ�. Remember to pay attention to the relation of force and μs�s. View Available Hint(s)for Part B Hint 1for Part B. Relation between frictional force and μs�s The force of friction, in this case, is not equal to μsN2�s�2. In general, Ffriction<μsFnormal�friction<�s�normal. Only when the object is on the verge of slipping does Ffriction=μsFnormal�friction=�s�normal. To find f� recall that the torques about the point at which the ladder touches the wall must be zero. Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value type f� =
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A uniform ladder with mass m2�2 and length L� rests against a smooth wall. (Figure 1)A do-it-yourself enthusiast of mass m1�1 stands on the ladder a distance d� from the bottom (measured along the ladder). The ladder makes an angle θ� with the ground. There is no friction between the wall and the ladder, but there is a frictional force of magnitude f� between the floor and the ladder. N1�1 is the magnitude of the normal force exerted by the wall on the ladder, and N2�2 is the magnitude of the normal force exerted by the ground on the ladder. Throughout the problem, consider counterclockwise torques to be positive. None of your answers should involve π� (i.e., simplify your trig functions).
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Part APart complete
Part BSuppose that the actual coefficent of friction is one and a half times as large as the value of μmin�min. That is, μs=(3/2)μmin�s=(3/2)�min. Under these circumstances, what is the magnitude of the
Express your answer in terms of m1�1, m2�2, d�, L�, g�, and θ�. Remember to pay attention to the relation of force and μs�s.
View Available Hint(s)for Part B
Hint 1for Part B. Relation between frictional force and μs�sThe force of friction, in this case, is not equal to μsN2�s�2. In general, Ffriction<μsFnormal�friction<�s�normal. Only when the object is on the verge of slipping does Ffriction=μsFnormal�friction=�s�normal. To find f� recall that the torques about the point at which the ladder touches the wall must be zero.
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