How Does Static Friction Affect Crate Stability on a Moving Lorry?

[FDhint]

A flatbed lorry travels at 100km/hr around a corner of radius 300m. inwardly banked at 10°
On the back of the lorry is a crate whose mass is 200kg. The cofficient of static fricition between the crate and the lorry being 0.7.

A) State the magnitude and direction (up or down the bed) of the frictional force F acting on the crate.

I am struggling with this question i believe you have to use the F= Mv^2/r to find the force but where does the static friction come into the equation?

Can anyone help please!

B) If the lorry were to enter a corner with a bank angle of 40° and radius of 150m. what is the range of speeds over which the lorry may travel without the crate slipping?

Does this question require the use of the SUVAT equations?

Can anyone help please!

Any working outs will be very useful

Thank you

 

[Simon Bridge]

Draw the free-body diagram (head on).
Draw all the forces acting on the load.

There must be a net unbalanced force pointing to the center of the turn equal to the centripetal force.

 

[CWatters]

where does the static friction come into the equation

What stops the load sliding off the truck as it goes around the corner?

mv²/r is just one of the forces acting on the load. For example what other forces would act on the load if the truck was stationary on the banked road?

As Simon said draw the diagram and work out all the forces.

 

[CWatters]

Part B is an variant of Part A. Won't need SUVAT equations.

Aside: Have you ever seen the banked track in a velodrome? The track is so steep that if a rider goes around too slowly they sometimes slide down the steep banking. On the other hand what would happen if you went around a banked track in a car way too fast or the banking wasn't steep enough? Under the right conditions there is a minimum and a maximum safe speed for banked tracks. Too slow and you slide down. Too high and you fly off the top. Your problem is similar except instead of friction between the tires and track being the issue is the friction between load and flat bed.

 

[rezeew]

A) The frictional force acting on the crate can be calculated using the formula F = μN, where μ is the coefficient of static friction and N is the normal force between the crate and the lorry. In this case, the normal force is equal to the weight of the crate, which can be calculated as mg, where m is the mass of the crate and g is the acceleration due to gravity. Therefore, the frictional force can be calculated as F = μmg. The direction of this force will be towards the center of the circular motion, which is inwardly towards the lorry's center.

B) This question does not require the use of SUVAT equations. To determine the range of speeds over which the lorry may travel without the crate slipping, we need to consider the maximum frictional force that can be exerted on the crate without it slipping. This maximum force can be calculated as μN, as explained in part A. The normal force, in this case, will be equal to the weight of the crate, which can be calculated as mg. Therefore, the maximum frictional force is μmg. We also know that the maximum frictional force is equal to the centripetal force required to keep the crate in circular motion, which can be calculated as mv^2/r, where v is the speed of the lorry and r is the radius of the curve. Therefore, we can set μmg = mv^2/r and solve for v to find the range of speeds over which the lorry can travel without the crate slipping.