A 2.5 kg block is initially at rest on a horizontal surface....

Name: Problem 5, Chapter 6: Principles of Physics
Uploaded: 2022-03-15T00:18:55-08:00
Duration: 5 min 14 s
Channel: Sandeep Kumar Dhania
Description: A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force of magnitude 6.0 N and a vertical force are then applied to the block (Fig. 6-15). The coefficients of friction for the block and surface are ms 0.40 and mk 0.25. Determine the magnitude of the frictional force acting on the block if the magnitude of is (a) 8.0 N, (b) 10 N, and (c) 12 N.

A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force of magnitude 6.0 N and a vertical force are then applied to the block (Fig. 6-15). The coefficients of friction for the block and surface are ms 0.40 and mk 0.25. Determine the magnitude of the frictional force acting on the block if the magnitude of is (a) 8.0 N, (b) 10 N, and (c) 12 N.

A 2.5 kg block is initially
at rest on a horizontal surface. A horizontal force of magnitude 6.0 N and a vertical force are
then applied to the block (Fig. 6-15).
The coefficients of friction for the
block and surface are ms 0.40 and
mk 0.25. Determine the magnitude
of the frictional force acting on
the block if the magnitude of is (a) 8.0 N, (b) 10 N, and (c) 12 N.

Key Concepts

Force Components

Decomposing forces into their horizontal and vertical components is a fundamental skill in mechanics. By resolving the applied forces into these components, one can analyze how vertical forces modify the normal force and, in turn, influence friction, while horizontal forces are directly related to initiating or sustaining motion.

Coefficients of Friction

Coefficients of friction, both static and kinetic, are dimensionless values that characterize the frictional interaction between two surfaces. The static friction coefficient determines the maximum frictional force available to prevent motion, while the kinetic friction coefficient applies when the object is sliding. These coefficients are essential for quantifying and comparing friction under different conditions.

Normal Force

The normal force is the contact force exerted by a surface perpendicular to the object resting on it. It is affected not only by the weight of the object due to gravity but also by any additional vertical forces applied to the object. In problems like this, correctly determining the normal force is crucial because it directly influences the magnitude of frictional forces acting on the object.

Frictional Force

Frictional force is the resistive force that opposes the relative motion between two surfaces in contact. It is calculated as the product of the coefficient of friction and the normal force. Understanding how friction behaves, whether it is static (preventing motion) or kinetic (opposing motion during movement), is key to solving dynamics problems involving surfaces.

Transcript

00:01 In this question, a box of 2 .5 kg, so the mass of the box is 2 .5 kg, is initially at rest on horizontal surface.

00:11 The horizontal force of magnitude f -dash, so f -dash, the value of f -dash is given as 6 newton, and a vertical force of p.

00:24 So the vertical force is along this direction, p.

00:28 The coefficient of friction is given for the static condition mu s is equals to 0 .4 and for the kinetic condition it is given as mu k is 2 .25 so the value of p is gradually varied about for the first condition it is 8 newton the value of p for the second condition is 10 newton and for the third condition it is 12 newton so we are required to find out the magnitude of frictional force acting on the block so we need to find out the value of small f so let us draw the normal reaction we know the frictional force is given by the formula f is equal to mu multiplied by n so we know n is equals to m g minus p m g in the downward direction and p along the upward direction so we get the normal reaction is equal to n g minus p therefore for the static condition f s is equal to mu s cross n which is equal to mu s multiplied by m g minus p so for the first condition the static friction is equal to 0 .4 multiplied by 2 .5 multiplied by 9 .8 minus 8.

02:17 And we get the value 6 .6 newton.

02:22 For the second condition, we get f s is equal to 0 .4 multiplied by 2 .5 multiplied by 9 .8 minus 12, which is equal to 5 .8 newton and for the third condition we get f s is equal to 0 .4 multiplied by 2 .5 multiplied by 9 .8 minus 12 which is equals to 5 neutrons so it is given the magnitude of the force in the horizontal direction f dash is is equals to 6 newton so for the first condition when p is equal to 8 newton we get 6 .6 newton which is greater than the magnitude of f dash but in the second and third condition when the value of p is 10 newton and 12 newton respectively it is less than the magnitude of the horizontal force so so at this condition the block is in more motion...

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