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Chapter 6, Problem 19P

Review. A window washer pulls a rubber squeegee down a very tall vertical window. The squeegee has mass 160 g and is mounted on the end of a light rod. The coefficient of kinetic friction between the squeegee and the dry glass is 0.900. The window washer presses it against the window with a force having a horizontal component of 4.00 N. (a) If she pulls the squeegee down the window at constant velocity, what vertical force component must she exert? (b) The window washer increases the downward force component by 25.0%, while all other forces remain the same. Find the squeegee’s acceleration in this situation. (c) The squeegee is moved into a wet portion of the window, where its motion is resisted by a fluid drag force R proportional to its velocity according to R = −20.0v, where R is in newtons and v is in meters per second. Find the terminal velocity that the squeegee approaches, assuming the window washer exerts the same force described in part (b).

(a)

Expert Solution
Check Mark
To determine

The vertical force component that washer must exert when she pulls the squeegee down the window at constant velocity.

Answer to Problem 19P

The vertical component of the force that washer must exert when she pulls the squeegee down the window at constant velocity is 2.03N .

Explanation of Solution

Given information:

The mass of the squeegee is 160g , the coefficient of kinetic friction between the squeegee and dry glass is 0.900 and the horizontal component of the force against the window is 4.00N .

Draw the diagram for the given condition.

Bundle: Physics For Scientists And Engineers With Modern Physics, Loose-leaf Version, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Single-term, Chapter 6, Problem 19P

Figure I

The horizontal component of the force exerted by washer against the window is given as,

FH=Fsinθ

  • F is the force exerted by washer against the window.
  • FH is the horizontal component of the force exerted by washer against the window.

The vertical component of the force exerted by washer against the window is given as,

FV=Fcosθ

  • FV is the vertical component of the force exerted by washer against the window.

The net vertical force for the given system is given as,

Verticalforce=FV+mgFs (I)

  • m is the mass of the squeegee.
  • g is the acceleration due to gravity.
  • Fs is the frictional force.

The expression for the frictional force is given as,

Fs=μFsinθ

  • μ is the coefficient of kinetic friction.

At constant velocity the net vertical force is zero.

Substitute μFsinθ for Fs in equation (I).

0=FV+mgμFsinθ

Rearrange the above expression for FV .

FV=μFsinθmg (II)

Substitute 0.900 for μ , 160g for m , 9.8m/s2 for g and 4.00N for Fsinθ in equation (II).

FV=(0.900)(4.00)(160g(1kg1000g))(9.8m/s2)=2.03N

Conclusion:

Therefore, the vertical component of the force that washer must exert when she pulls the squeegee down the window at constant velocity is 2.03N .

(b)

Expert Solution
Check Mark
To determine

The acceleration of the squeegee if the washer increases the downward force component by 25% .

Answer to Problem 19P

The acceleration of the squeegee when the washer increases the downward force component by 25% is 3.18m/s2 .

Explanation of Solution

Section 1:

To determine: The net vertical force when the downward force component is increased by 25% .

Answer: The net vertical force when the downward force component is increased by 25% is 0.5055N .

Given information:

The mass of the squeegee is 160g , the coefficient of kinetic friction between the squeegee and dry glass is 0.900 and the horizontal component of the force against the window is 4.00N .

The new vertical component of the force exerted by washer against the window is given as,

FV=1.25FV

The net vertical force is given as,

FVn=FV+mgμFH

Substitute 1.25FV for FV in above equation.

FVn=1.25(FV)+mgμ(FH) (III)

Substitute 4.00N for FH , 160g for m , 0.900 for μ , 9.8m/s2 for g and 2.03N for FV in equation (III).

FVn=1.25(2.03N)+(160g(1kg1000g))(9.8m/s2)(0.900)4.00N=0.5055N

Section 2:

To determine: The acceleration of the squeegee.

Answer: The acceleration of the squeegee is 3.18m/s2 .

Given information:

The mass of the squeegee is 160g , the coefficient of kinetic friction between the squeegee and dry glass is 0.900 and the horizontal component of the force against the window is 4.00N .

Formula to calculate the force is,

FVn=ma

  • a is the acceleration.

Rearrange the above expression for a .

a=FVnm (IV)

Substitute 160g for m and 0.5055N for FVn in equation (IV).

a=0.5055N(160g(1kg1000g))=3.18m/s2

Conclusion:

Therefore, the acceleration of the squeegee when the washer increases the downward force component by 25% is 3.18m/s2 .

(c)

Expert Solution
Check Mark
To determine

The terminal velocity that the squeegee approaches.

Answer to Problem 19P

The terminal velocity that squeegee approaches is 0.205m/s .

Explanation of Solution

Given information:

The mass of the squeegee is 160g , the coefficient of kinetic friction between the squeegee and dry glass is 0.900 , the horizontal component of the force against the window is 4.00N and the fluid drag force is 20.0v .

According to the given condition,

R=(FV+mg)

Substitute 20.0v for R and 1.25(FV) in above expression.

20.0v=(1.25(FV)+mg)

Rearrange the above expression for v .

v=((1.25(FV)+mg))20.0 (V)

Substitute 160g for m , 9.8m/s2 for g and 2.03N for FV in equation (V).

v=(1.25(2.03N)+(160g(1kg1000g))9.8m/s2)20.0=0.205m/s

Conclusion:

Therefore, the terminal velocity that squeegee approaches is 0.205m/s .

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Chapter 6 Solutions

Bundle: Physics For Scientists And Engineers With Modern Physics, Loose-leaf Version, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Single-term

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