A 4000 kg truck is parked on a 7.0° slope. How big is the friction force on the truck?

1 Answer
Morgan
Jan 4, 2017

#vecf_s~~4.8xx10^3 N#

Explanation:

Here is a basic force diagram for an object on an inclined plane:

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Note that the angle is certainly not to scale.
Note that I will define down the ramp as the positive direction.

We can determine the frictional force acting on the truck with statements of the parallel and perpendicular components of the net force acting on the truck.

#sumF_x=vecF_(gx)-vecf_s=mveca_x#

#sumF_y=vecn-vecF_(gy)=mveca_y#

To prevent the truck from rolling down the incline, we want a horizontal acceleration of zero. The vertical acceleration should also be zero, as we certainly don't want the truck to be moving in the vertical direction. This leaves us with a net force of zero on the truck. Note this is static friction.

#sumF_x=vecF_(gx)-vecf_s=0#

#sumF_y=vecn-vecF_(gy)=0#

We can solve for the force of friction using the parallel sum of forces statement. The perpendicular forces are not necessary.

#vecF_(gx)-vecf_s=0#

#=>vecF_(gx)=vecf_s#

The parallel component of the force of gravity is given as #mgsin(theta)# for this situation, which we can see from the force diagram. I will explain this at the end if it is unclear.

#=>vecf_s=mgsin(theta)#

Using our known values...

#=>vecf_s=(4000kg)(9.8m/s^2)sin(7^o)#

#=>vecf_s~~4777N#

You might write #4.8xx10^3N# where significant figures are concerned.

Explanation of #vecF_(gx)#:

By the laws of geometry, the angle between #vecF_g# and #vecF_(gy)# is equal to the angle of the incline. We see that #vecF_(gx)# is opposite that angle. Additionally, the force of gravity is given by #vecF_g=mg#, which serves as the hypotenuse of the triangle formed.

#sin(theta)=(op.)/(hyp.)#

#=>sin(theta)=(vecF_(gx))/(vecF_g)#

#=>vecF_(gx)=vecF_g*sin(theta)#

#=>vecF_(gx)=mgsin(theta)#

Tip: I've seen problems which ask the exact same question, but also provide the coefficient of static friction. This makes you think you should use

#vecf_(smax)=mu_svecn#

The problem is that the truck should not be at maximum static friction. I would not want to be near that truck! It would not take much to send it rolling down the hill. Unless specifically asked for the maximum static friction, use the above method.

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