Question

A 3600 kg truck is parked on a 14 ∘ slope. You may want to review (Pages 141 - 145) . Part A How big is the friction force on the truck? The coefficient of static friction between the tires and the road is 0.90. Express your answer to two significant figures and include the appropriate units.

Answer 1

The friction force on a 3600 kg truck parked on a 14-degree slope with a coefficient of static friction of 0.90 is approximately 30648 N.

Step-by-step explanation:

The question is asking to calculate the friction force on a parked truck on a slope using the given coefficient of static friction. First, we find the component of the truck's weight that is parallel to the slope, which is calculated by multiplying the truck's weight by the sine of the slope's angle. The weight of the truck (W) equals its mass (m) times the acceleration due to gravity (g). The parallel component (Wparallel) is given by W * sin(θ). The friction force (fs) is then calculated by multiplying the normal force (N), which is equal to the component of the truck's weight perpendicular to the slope (W * cos(θ)), by the coefficient of static friction (μs).

I've computed the necessary values:

• The weight of the truck (W) is 3600 kg × 9.8 m/s2 = 35280 N.
• Component of weight parallel to slope: Wparallel = 35280 N × sin(14 degrees) ≈ 8503 N.
• Component of weight perpendicular to slope: Wperpendicular = 35280 N × cos(14 degrees) ≈ 34053 N.
• Friction force: fs = μs × N = 0.90 × 34053 N ≈ 30648 N.

Therefore, the friction force on the truck is approximately 30648 N, expressed to two significant figures.

Answer 2

Answer:8535 N

Step-by-step explanation:

Given

mass of truck
`(m_t)`=3600 kg

Truck is parked on a
`14 ^(\circ)` slope

so its sin component of weight will try to pull it down the slope and friction is resist its motion

Weight sin component
`=m_t\cdot gsin\theta =3600* 9.8* sin(14)`

=8535 N

Maximum Frictional force
`=\mu N`

`N=mg\cos \theta`

`f_r=0.9* 3600* 9.8* \cos (14)=30,808.829 N`

so friction Force adjust itself to imparts 8535 N frictional force.