A toy chest and its contents have a combined weight of 200...

A toy chest and its contents have a combined weight of 200 N. The coefficient of static friction between toy chest and floor is 0.47. The child in Fig. 6-36 attempts to move the chest across the floor by pulling on an attached rope. (a) If u is 42°, what is the magnitude of the force that the child must exert on the rope to put the chest on the verge of moving? (b) Write an expression for the magnitude F required to put the chest on the verge of moving as a function of the angle u. Determine (c) the value of u for which F is a minimum and (d) that minimum magnitude.

A toy chest and its contents
have a combined weight of 200 N. The coefficient of static friction
between toy chest and floor is 0.47. The child in Fig. 6-36 attempts
to move the chest across the floor by pulling on an attached rope.
(a) If u is 42°, what is the magnitude of the force that the child
must exert on the rope to put the chest on the verge of moving? (b)
Write an expression for the magnitude F required to put the chest
on the verge of moving as a function of the angle u. Determine (c)
the value of u for which F is a minimum and (d) that minimum
magnitude.

Key Concepts

Static Friction

Static friction is the force that resists the initiation of motion between two surfaces in contact. It acts up to a maximum value that is proportional to the normal force through the coefficient of static friction, and its threshold value determines when an object will begin to move.

Coefficient of Static Friction

The coefficient of static friction is a dimensionless number representing the ratio of the maximum static friction force to the normal force between two surfaces. It is a key parameter in calculating whether a given applied force is sufficient to initiate movement.

Force Resolution and Free Body Diagrams

Force resolution involves breaking a force into its horizontal and vertical components using trigonometric functions. This process is crucial for setting up free body diagrams, which help in visualizing and analyzing the forces acting on an object in equilibrium or at the verge of motion.

Equilibrium Conditions

Equilibrium conditions require that the sum of forces in each direction is zero. When an object is on the verge of moving, the applied force, its components, and the frictional and gravitational forces are balanced, allowing for the determination of the threshold force required to trigger motion.

Optimization in Physics

Optimization in physics involves finding the conditions that minimize or maximize a certain quantity, such as the applied force required to overcome static friction. This process typically requires setting up an equation for the force as a function of a variable (like the angle of application) and then finding the value that minimizes the function, often using calculus.

Transcript

00:01 In this question, a toy chest and its contents have a combined weight of w is equal to 200 newton.

00:09 A coefficient of static friction within the toy chest and the floor is equals to 0 .47.

00:19 So a child attempts to move the chest across the floor by pulling on an attached rope.

00:28 So this is the direction of the attached stroke.

00:31 The direction of applied force which is at an angle 42 degree with respect to the horizontal plan so you are asked for the first part the magnitude of force f -dash for which the box will start to move so first we will check the force acting on the object so this is the direction of the force f -dash so this is the direction of a itself a cost component which is f dash cost theta where theta is equal to 42 degrees and this is the sign component which is f dash sine theta so this is the direction of normal and this is the direction of weight which is given as w and we know weight of an object is given by mass multiplied by the acceleration duty gravity which is n g and this is the direction of the friction force which is f s so from here we get the normal reaction fn is equal to w minus the sine component of f dash so this is f dash sine theta we know frictional force is given as for the static friction f s is equal to mu static multiplied by the normal reaction which is equal to mu s multiplied by w minus f dash sine theta for the box to just start moving f s must be zero how is that possible we will apply a force of equal magnitude on the opposite direction to that of f s so in this figure we can see the horizontal force opposite to the f s is nothing but the cost component of the force f dash which is f dash cos phi so we can write f dash cos phi must be equal to fs for the box to just start moving now putting the value f s we can write f dash cos phi is equal to mu s multiplied by w minus dash sine theta now putting the values we can write f dash cost theta plus mu as f dash sine theta is equals to mu s w or f dash cost 42 degree plus 0 .47 multiplied by sine 42 degree is equal to 0 .47 multiplied by the weight of the crate which is equal to 200 newton or from here we get the value of f dash is equal to 88 .87 newton now for the second part of the question we are asked to write down the expression for the magnitude of f required to put the chest on the verge of moving with respect to the function of angle theta in the previous part of the question we have got the equation if cost theta is equal to fs so using this equation we will find out the dependency of angle theta or we can write f dash cos theta is equals to nothing but mu is multiplied by w minus f dash sine theta from here we can write f dash in common cos theta plus mu sine theta is equal to mu s multiplied by m g or we can write f dash is equals to mu s multiplied by m multiplied by g divided by cost theta plus mu as sine theta so this is how the angle theta results to the change in the applied force in the third part of the question we are asked to find out the angle theta at which f will be minimum therefore at f minimum we can write df by d theta is equal to 0...

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