Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 8, Problem 124P

(a)

To determine

An algebraic solution for d in terms of M,m,μs,L and θ.

(a)

Expert Solution
Check Mark

Answer to Problem 124P

An algebraic solution for d in terms of M,m,μs,L and θ is d=(μsM+mMtanθm2M)L.

Explanation of Solution

The system is in equilibrium until the ladder begins to slip.

Write the expression for the Newton’s second law for the equilibrium.

ΣFx=0 (I)

Here, ΣFx is the net external force acting on the system in the x direction.

Write the expression for ΣFx.

ΣFx=Nwf

Here, Nw is the normal force acting on the ladder due to the wall and f is the frictional force acting on the base of the ladder due to the floor

Put the above equation in equation (I).

Nwf=0Nw=f (II)

The frictional force has its maximum possible magnitude at the person’s highest point.

Write the equation for the maximum possible magnitude of frictional force.

f=μsNf

Here, μs is the coefficient of static friction between the floor and the ladder and Nf is the normal force on the ladder due to the floor

Put the above equation in equation (II).

Nw=μsNf (III)

The net external force on the system in the y direction must be zero.

ΣFy=0 (IV)

Here, ΣFy is the net external force on the system in the y direction

Write the expression for ΣFy.

ΣFy=NfMgmg

Here, M is the mass of the person, g is the acceleration due to gravity and m is the mass of the ladder

Put the above equation in equation (IV) and rewrite it for Nf.

NfMgmg=0Nf=Mg+mg=(M+m)g

Put the above equation in equation (III).

Nw=μs(M+m)g (V)

Choose the axis of rotation at the contact point between the ladder and the floor.

The net torque about the rotation axis is zero in equilibrium.

Στ=0 (VI)

Here, Στ is the net external torque on the system

Write the expression for Στ.

Στ=Nf(0)+f(0)NwLsinθ+mg(12Lcosθ)+Mgdcosθ=NwLsinθ+mg(12Lcosθ)+Mgdcosθ

Here, L is the length of the ladder, θ is the angle between the ladder and the floor and d is the distance from the bottom of the ladder to the point where the person stands

Put the above equation in equation (VI) and rewrite it for d.

NwLsinθ+mg(12Lcosθ)+Mgdcosθ=0d=NwLsinθ12mgLcosθMgcosθ

Put equation (V) in the above equation.

d=μs(M+m)gLsinθ12mgLcosθMgcosθ=[μs(M+m)sinθm2cosθ]LMcosθ=μs(M+m)sinθMcosθ+m2cosθMcosθ=(μsM+mMtanθm2M)L (VII)

Conclusion:

Therefore, the algebraic solution for d in terms of M,m,μs,L and θ is d=(μsM+mMtanθm2M)L.

(b)

To determine

Use the expression derived in part (a) to show that placing the ladder at a larger angle θ enables the person to climb farther up the ladder without having it slip, all other things being equal.

(b)

Expert Solution
Check Mark

Answer to Problem 124P

It is shown using the expression derived in part (a) that placing the ladder at a larger angle θ enables the person to climb farther up the ladder without having it slip, all other things being equal.

Explanation of Solution

Refer to equation (VII) derived in part (a).

The value of d increases with increase in tanθ according to equation (VII). The value of tanθ increases on the interval 0θ90°. This implies placing the ladder at a larger angle (that is more nearly vertical) allows the person to climb farther up the ladder without having it slip.

Conclusion:

Thus, it is shown using the expression derived in part (a) that placing the ladder at a larger angle θ enables the person to climb farther up the ladder without having it slip, all other things being equal.

(c)

To determine

The maximum angle that enables the person to climb all the way to the top of the ladder.

(c)

Expert Solution
Check Mark

Answer to Problem 124P

The maximum angle that enables the person to climb all the way to the top of the ladder is 63°.

Explanation of Solution

To climb up to the top of the ladder, set the value of d to the length of the ladder.

d=L

Put the above equation in equation (VII).

L=(μsM+mMtanθm2M)LL+L(m2M)=(μsM+mMtanθ)L1+m2M=μsM+mMtanθ2M+m2=μs(M+m)tanθtanθ=2M+m2μs(M+m)

Rewrite the above equation for θ.

θ=tan12M+m2μs(M+m) (VIII)

Conclusion:

Given that the value of μs is 0.45 , mass of the person is 60.0 kg and the  mass of the ladder is 15.0 kg.

Substitute 60.0 kg for M , 15.0 kg for m and 0.45 for μs in equation (VIII) to find the value of θ.

θ=tan12(60.0 kg)+15.0 kg2(0.45)(60.0 kg+15.0 kg)=63°

Therefore, the maximum angle that enables the person to climb all the way to the top of the ladder is 63°.

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Chapter 8 Solutions

Physics

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