A worker pushes horizontally on a 35 kg crate with a force of magnitude 110 N . The coefficient

A worker pushes horizontally on a 35 kg crate with a force...

A worker pushes horizontally on a 35 $\mathrm{kg}$ crate with a force of magnitude 110 $\mathrm{N}$ . The coefficient of static friction between the crate and the floor is 0.37 (a) What is the value of $f_{x, \max }$ under the circumstances? (b) Does the crate move? (c) What is the frictional force on the crate from the floor? (d) Suppose, next, that a second worker pulls directly upward on the crate to help out. What is the least vertical pull that will allow the first worker's 110 $\mathrm{N}$ push to move the crate? (e) If, instead, the second worker pulls horizontally to help out, what is the least pull that will get the crate moving?

A worker pushes horizontally on a 35 $\mathrm{kg}$ crate with a force of magnitude 110 $\mathrm{N}$ . The coefficient of static friction between the crate and the floor is 0.37 (a) What is the value of $f_{x, \max }$ under the circumstances? (b) Does the crate move? (c) What is the frictional force on the crate from the floor? (d) Suppose, next, that a second worker pulls directly upward on the crate to help out.
What is the least vertical pull that will allow the first worker's 110 $\mathrm{N}$ push to move the crate? (e) If, instead, the second worker pulls horizontally to help out, what is the least pull that will get the crate moving?

Key Concepts

Static Friction

Static friction is the force that resists the initiation of sliding motion between two surfaces in contact. It is characterized by a maximum value, defined by the product of the coefficient of static friction and the normal force, beyond which motion begins. This concept is crucial when determining if a given applied force can overcome resistance and set an object in motion.

Normal Force

The normal force is the perpendicular force exerted by a surface to support the weight of an object resting on it. It is determined by the object's weight and any additional vertical forces acting on the object. In problems involving friction, the normal force plays a key role because it directly affects the maximum static friction available.

Force Equilibrium

Force equilibrium involves analyzing the sum of forces acting on an object to determine if the object remains stationary or moves. When all forces balance out, the object stays at rest. This concept is used to assess whether an applied force is sufficient to overcome friction and initiate movement.

Free-Body Diagram Analysis

A free-body diagram is a graphical representation that depicts all the forces acting on an object. It helps in visualizing forces such as applied forces, friction, gravitational force, and normal force, and is essential for setting up the equations used to solve for unknowns in dynamics problems.

Effect of Vertical Assistance on Friction

When an additional vertical force is applied (such as pulling upward), it impacts the normal force by reducing or increasing it, thereby altering the maximum static friction. Understanding this effect is important in scenarios where modifying the frictional resistance is desired, allowing a smaller horizontal force to overcome static friction and initiate movement.

Transcript

00:01 So we're going to denote the force exerted by the worker on the crate as f equaling 110 newton's.

00:08 And we can say that then the maximum static frictional force would be equaling the coefficient of static friction multiplied by the force normal.

00:17 For part a, when we apply newton's second law here in the vertical direction, we can say that the force normal would be equal to the gravitational force mg.

00:25 And so we can say that the maximum static frictional force would then be equaling.

00:30 To the coefficient of static friction mg.

00:34 This is going to be equalling to 0 .37, multiplied by 35 kilograms, multiplied by the acceleration due to gravity, 9 .8 meters per second squared.

00:45 And we find that then this is equaling approximately 130 newtons, and we can then say that the maximum static frictional force is greater than f.

00:58 For part b, we can see that the block is initially at rest and it stays at rest because the force f is less than the maximum static frictional force.

01:11 If the force can't exceed that maximum static frictional force, then the block will stay at rest.

01:18 And we can say again, block stays at rest.

01:26 And for part c, we can say that applying newton's second law here to the horizontal direction, we can say that the magnitude of the static frictional force exerted here on the crate would be equaling f equaling 110 newton's.

01:44 So that would be your answer for part c, and then for part d, we have that.

01:49 First, we're going to say the upward force exerted by the second worker as f sub 2, exerted by second worker.

02:11 And given this, we can apply, again, newton's second law...

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