Question

A 5000 kg truck is parked on a 7.0? slope. How big is the friction force on the truck?

Express your answer to two significant figures and include the appropriate units.

fs=

Answer 1

Concept and reason

The concept of forces acting on an object at elevation is used to find the magnitude of frictional force.

The formula for the normal force and the weight is used to find the frictional forces.

Fundamentals

According to the relation between forces, the force due to gravity is balanced by the normal force acting vertically upward can canceling the weight.

The force due to gravity acts on the object below the body and the force due to normal acts on the body due the object below. Also, the magnitude of the both forces is equal in case the body is at rest.

When a body is placed, and inclined plane and the body is at rest the frictional force is canceling all other forces to balance the static motion.

The force due to gravity acts on the object below the body and the force due to normal acts on the body due the object below.

The object is at inclined plane and the weight of the body always points toward downwards. Due to this horizontal and vertical component of weight are required so that one component points towards the base of body and other directs towards the downside of the inclination.

From this it can be clearly observed in the image that vertical component of weight as ${W_y} = W{\rm{Sin}}7^\circ$ and the horizontal component of weight as

${W_x} = W{\rm{Cos}}7^\circ$ .

Here the normal force cancels out the ${W_x} = W{\rm{Cos}}7$ in magnitude and is opposite in the direction.

As the truck is stationary or is parked hence the static frictional fore is acting against the forces directing towards the down side of inclination and is opposite in the direction.

From this it can be clearly observed in the image that vertical component of weight as ${W_y} = W{\rm{Sin}}7$ and the horizontal component of weight as

${W_x} = W{\rm{Cos}}7^\circ$ .

The component of weight ${W_x}$ is canceled out in magnitude by the normal force that is opposite in the direction. The other component of weight is only parameter that can be compared with static frictional force.

Here the static frictional force is equal to the ${W_y} = W{\rm{Sin}}7$ in magnitude and is opposite in the direction.

Let force due to static friction is denoted by ${F_s}$ .

Hence

${F_s} = {W_y}$

${F_s} = W{\rm{Sin}}7^\circ$ .

As the weight of the truck is given by the expression $W = mg$ , where m is the mass and g is the gravity of the earth.

It is given that, mass $m = 5000{\rm{kg}}$ and gravity $g = 9.8{\rm{m}}{{\rm{s}}^{ - 2}}$ .

Now substituting the values,

$\begin{array}{c}\\W = 5000{\rm{kg}} \times 9.8{\rm{m}}{{\rm{s}}^{ - 2}}\\\\ = 49000\;{\rm{N}}\\\end{array}$

Then the weight of the truck is $W = 49000{\rm{N}}$ .

Now from expression is given as follows:

$\begin{array}{c}\\{F_s} = W{\rm{Sin}}7^\circ \\\\ = 49000\sin 7^\circ \\\\ = 5971.59{\rm{ N}}\\\end{array}$

The magnitude of the force due to the static friction is ${F_s} = 5971.59{\rm{N}}$ .

Ans:

The magnitude of the force due to the static friction is ${F_s} = 5971.59{\rm{N}}$ .