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Answered

A 4000kg truck is parked on a 15.0∘ slope.


How big is the friction force on the truck?

Verified Answer :

MissPhiladelphia

Since the truck is not moving, the frictional force must be at least as large as the parallel component of the truck's weight (gravitational force). 
The parallel component of the gravitational force is 
F= m g sin(15) 
therefore the frictional force is at least
40009.8sin(15) =
The answer would be

10,153 N
I hope my answer helped you.

Answer:

Frictional force, F = 10145.70 N

Explanation:

It is given that,

Mass of the truck, m = 4000 kg

Angle between ground and the truck, \(\theta=15\)

The force that opposes the motion of abject is called frictional force. Here, the truck is at rest. So, the frictional force is given by :

\(F=N\ sin\theta\)

N is the normal force, N = mg

\(F=mg\ sin\theta\)

\(F=4000\ kg\times 9.8\ m/s^2\ sin(15)\)

F = 10145.70 N

So, the frictional force on the truck is 10145.70 N. Hence, this is the required solution.

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