Question

A tractor drags a 313kg plow across a field at 2.0 m/s . If the coefficient of friction between the plow and the ground is 0.570, how much force does the tractor apply to the plow ?

Answer 1

Answer:

The tractor applies 1,748.418 N to the plow.

Explanation:

Net Force

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

Fn=m.a

If the acceleration is zero, then the net force is also zero. That means all forces acting on the mass are balanced.

The tractor drags the m=313 Kg plow at a constant speed, thus the acceleration is zero and so is the net force.

The horizontal forces acting on the plow are:

• The force applied by the tractor Ft
• The friction of the ground Fr

Since both forces are balanced, then:

Ft=Fr

The friction force is calculated as:

$Fr=\mu \cdot N$

Where $\mu$ is the coefficient of friction between the plow and the ground and N is the normal force. In the situation described, the normal force is equal to the weight of the plow:

$N=W=m.g=313*9.8=3,067.4$

The normal force is N=3,067.4 N

The friction force is now calculated:

$Fr=0.579 \cdot 3,067.4\ N$

$Fr=1,748.418\ N$

Thus, the tractor applies 1,748.418 N to the plow.