COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Question
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Chapter 2, Problem 76P

(a)

To determine

The normal force applied on the crate by the ramp.

(a)

Expert Solution
Check Mark

Answer to Problem 76P

The normal force applied on the crate by the ramp is 75.2N

Explanation of Solution

The weight of the crate is 80.0N. The angle of inclination of the ramp is 20.0°. Consider the ramp to be x axis.

The following pic shows the FBD of the system.

The net force along the y-axis is zero as there is no motion in y-axis.

  Nmgcosθ=0                                                                

Here, N is the normal force, m is the mass, g is the acceleration due to gravity, and θ is the angle of inclination of the ramp.

Re-write the above expression to get an expression for N.

  N=mgcosθ        (I)

Conclusion:

Substitute 80.0N for mg, 20.0° for θ in equation (I)

  N=(80.0N)cos(20.0°)=75.2N

The normal force applied on the crate by the ramp is 75.2N

(b)

To determine

The magnitude and direction of the interaction partner of the normal force.

(b)

Expert Solution
Check Mark

Answer to Problem 76P

The interaction partner of the normal force will have magnitude of 75.2N and applied in opposite direction to the normal force.

The normal force on the crate has magnitude of 75.2N. Thus the interaction partner of the normal force will have magnitude of 75.2N and applied in opposite direction to the normal force.

Explanation of Solution

According to the Newton’s third law the interaction partners have equal magnitude and opposite.

The normal force on the crate has magnitude of 75.2N. Thus the interaction partner of the normal force will have magnitude of 75.2N and applied in opposite direction to the normal force.

Conclusion:

The interaction partner of the normal force will have magnitude of 75.2N and applied in opposite direction to the normal force.

(c)

To determine

The static friction force between the crate and the ramp.

(c)

Expert Solution
Check Mark

Answer to Problem 76P

The static friction force between the crate and the ramp is 27.4N

Explanation of Solution

The weight of crate is 80.0N. The angle of inclination of the ramp is 20.0°. Consider the ramp to be  x-axis.

The following pic shows the FBD of the system.

When the crate is static, the force along the x-axis is zero.

  fsmgsinθ=0        (I)

Here, fs is the friction force, m is the mass, g is the acceleration due to gravity, θ is the angle of inclination of the ramp.

Re-write the above expression to get an expression for fs.

  fs=mgsinθ        (II)

Conclusion:

Substitute 80.0N for mg, 20.0° for θ in equation (I)

  fs=(80.0N)sin(20.0°)=27.4N

The static friction force between the crate and the ramp is 27.4N

(d)

To determine

The minimum possible value for the coefficient of static friction.

(d)

Expert Solution
Check Mark

Answer to Problem 76P

The minimum possible value for the coefficient of static friction is 0.364

Explanation of Solution

 The following pic show the FBD of the system,

The minimum coefficient of static friction is just enough to make the friction force equalize the component of weight in the direction of the crate’s motion.

  fs=mgsinθ        (I)

Here, fs is the friction force, m is the mass, g is the acceleration due to gravity, θ is the angle of inclination of the ramp.

Write the formal for the coefficient of static friction.

  μs=fsN        (II)

Here, μs is the coefficient of static friction, N is the normal force.

Write the formula for the normal force.

  Nmgcosθ=0        (III)

Substitute expression (I) and (III) in expression (II),

  μs=mgsinθmgcosθ=tanθ        (IV)

Conclusion:

Substitute  20.0° for θ in equation (IV)

  μs=tan(20.0°)=0.364

The minimum possible value for the coefficient of static friction is 0.364

(e)

To determine

The magnitude and direction of the contact force exerted on the crate by the ramp.

(e)

Expert Solution
Check Mark

Answer to Problem 76P

The magnitude of the contact force is 80.0N and it makes 70.0° with the x axis

Explanation of Solution

The weight of the crate is 80.0N. The angle of inclination of the ramp is 20.0°. Consider the ramp to be x axis.

The friction force and the normal force are the component of the contact force.

Write the formula for the magnitude of the contact force.

  |F|=fs2+N2        (I)

Here F is the contact force, fs is the friction force, N is the normal force.

Write the direction of the contact force.

  θ=tan1(Nfs)        (II)

Here, θ is the angle made by the contact force with the positive x axis.

Conclusion:

Substitute 75.2N for N,  27.4N for fs in equation (I)

  |F|=(27.4N)2+(75.2N)2=80.0N

Substitute 75.2N for N27.4N for fs in equation (I)

  θ=tan1(75.2N27.4N)=70.0°

The magnitude of the contact force is 80.0N and it makes 70.0° with x axis.

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Chapter 2 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

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