Figure 6-32 shows three crates being pushed over a concrete...

Figure 6-32 shows three crates being pushed over a concrete floor by a horizontal force of magnitude 425 N. The masses of the crates are m1 30.0 kg, m2 10.0 kg, and m3 20.0 kg. The coefficient of kinetic friction between the floor and each of the crates is 0.700. (a) What is the magnitude F32 of the force on crate 3 from crate 2? (b) If the crates then slide onto a polished floor, where the coefficient of kinetic friction is less than 0.700, is magnitude F32 more than, less than, or the same as it was when the coefficient was 0.700?

Figure 6-32 shows three crates
being pushed over a concrete floor
by a horizontal force of magnitude 425 N. The masses of the
crates are m1 30.0 kg, m2 10.0
kg, and m3 20.0 kg. The coefficient of kinetic friction between the
floor and each of the crates is 0.700.
(a) What is the magnitude F32 of the
force on crate 3 from crate 2? (b) If
the crates then slide onto a polished
floor, where the coefficient of kinetic friction is less than 0.700, is magnitude F32 more than, less than, or the same as it was when the
coefficient was 0.700?

Key Concepts

Contact Forces Between Objects

When multiple objects are in contact and a force is applied to one end, internal forces (or contact forces) develop between the objects. These forces ensure that all objects move with the same acceleration. Determining the magnitude of these forces often involves applying Newton’s Second Law to individual bodies within the system.

Free-Body Diagrams

Free-body diagrams are visual representations that isolate an object and depict all the forces acting upon it. They are an essential tool in mechanics for setting up the equations of motion for single bodies or systems, particularly in problems involving friction, applied forces, and contact forces between bodies.

Newton's Second Law

This fundamental principle states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). In problems involving multiple bodies in contact, each body’s acceleration is determined by the net force acting on the entire system, and this acceleration is used to relate the forces acting on individual bodies.

Kinetic Friction

Kinetic friction is the resistive force that opposes the motion of an object sliding over a surface. It is calculated using the formula f_k = ?_kN, where ?_k is the coefficient of kinetic friction and N is the normal force. The magnitude of this frictional force affects the net force and acceleration of objects in motion.

Transcript

00:01 In this question it is given that the values of m1 is equal to 30 kg m2 is equal to 10 kg and m3 is equal to 20 kg so a force of f dash is equal to 425 newton is applied to the system and the coefficient of friction between the surface of the and the floor is given by mu k is equal to 0 .7 so from here we can find the acceleration of the system so for that we need to find out the net force acting on the system so f net is given by f dash or the force applied minus the friction on loss or f s is the magnitude of friction and force so we know f s is equals to mu k times the normal reaction and the normal reaction is nothing but the total mass of the system which can be emitted by big m multiplied by the gravitation constant multiplied by the acceleration due to gravity which is g so here m is equals to m1 plus m2 plus m 3 which is equals to 30 plus 10 plus 20 which is equals to 60 kgs so from here we get n is equal to 60 g or f s is equal to 0 .7 multiplied by 60 g or we get f net equals to 425 minus 0 .7 multiplied by 60 g now according to newton's second law we can write f net is equal to the mass of the system which is m multiplied by the acceleration gained by the system or we can write 60 a is equal to 425 minus 0 .7 multiplied by the acceleration gained by the system or we can write 60 a is equal to 425 minus 0 .7 multiplied by 60 multiplied by 9 .8 or from here upon solving we get a is equal to 0 .2 to 2 3 meter per second square now for the first part of the question we are asked to find out the magnitude f2 3 of the force on the crate 3 from grade 2 so we have to find out the amount of force which the crate m2 applies on crate m3 so we know the mass of the third crate m3 is equal to 20 kgis let us draw the freebody diagram of the third crate which is m3 so this is the direction of the normal reaction this is the direction of the weight which is m3g this is the direction of the force f 3 2 and this is the direction of the force of acceleration which is f s 3 suppose so from here we get the normal reaction for the third crate is m 3g or f s 3 or the friction force is equals to mu k multiplied by n which is m 3g therefore f net for this is equal to f3 2 minus the frictional force which is f s 3 or from here we get f32 is equal to fnett plus fs 3 also according to the newton's second love we can write fnett is equal to f3 multiplied by the acceleration of the system which is this one or by putting the values we get f3 2 is equal to m 3a plus mu k m 3g which is equal to m3a plus mu k g by putting the values again 20 multiplied by 0 .223 plus 0 .7 multiplied by 9 .8 or the magnitude of the force exerted by the second crate on the third crate is equals to 1 for 1 .66 newton now in the second part of the question we are asked that whether the contact force between the second crate and the third crate varies with the coefficient of friction so let us so let us find out the magnitudes which are indirect dependent on the coefficient of friction.

06:18 So from these equation we can see the acceleration depends on the coefficient of friction.

06:29 Similarly here also we can see the contact force depends on the coefficient of friction.

06:38 But when the coefficient of friction increases or decreases, the acceleration also increases or decreases...

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