When the tire doesn't slip, it means the point of contact with the ground is at rest with respect to the ground. So, static friction acts
Max acceleration = μs *g = 0.37*9.8 = 3.626 m/s^2
(b)When tire slips, it means the point of contact with the ground is in motion and so kinetic friction acts.
Constant acceleration = μk * g = 2.058 m/s^2
Given
Mass of crate, m = 45 kg
coefficient of static friction,µs =0.370
coefficient of kinetic friction,µk = 0.210
a)The net force on the truck is
F = µs m g
ma = µs m g
a = 0.370 *9.8 m/s^2
a = 3.62600 m/s^2
Thus, the acceleration of the truck is 3.626 m/s^2
b) The net force on the crate is
F = µk m g
m a = µk m g
a = 0.210 *9.8 m/s^2
a = 2.058 m/s^2
Thus the acceleration of the crate relative to ground is 2.058 m/s^2
General Physics with Calc 1 8. A package of dishes (mass 50.0 kg) sits on the flatbed of a pickup truck with an open tailgate. The coefficient of static friction between the package and the truck's flatbed is 0.310, and the coefficient of kinetic friction is 0.240. (a) The truck accelerates forward on level ground. What is the maximum acceleration the truck can have so that the package does not slide relative to the truck bed? (b) The truck barely...
The flatbed truck starts from rest on a road whose constant radius of curvature is 38 m and whose bank angle is 8° Ifthe constant forward acceleration of the truck is 2.5 m/sec2, determine the smallest time t after the start of motion at which the crate on the bed begins to slide. The coefficient of static friction between the crate and truck bed is μ.-11 , and the truck motion occurs in a horizontal plane sec
Q5: A flatbed truck starts from rest on a road whose constant radius of curvature is 30 m and whose bank angle is 1°. If the constant forward acceleration of the truck is 2 m/s, determine the time t after the start of motion at which the crate on the bed begins to slide. The coefficient of static friction between the crate and truck bed is -0.3, and the truck motion occurs in a horizontal plane. (5.58 s 10°
Problem 9. You push on a 30kg object with a force of 200 N, but it refuses to move. The coefficient of static friction between the object and the floor is 0.80, what is the magnitude of the friction force on the object? Solution El = 200 N Problem 10 A flatbed truck slowly tilts its bed upward to dispose of a 95 kg box. When the tilt angle just exceeds 23°, the crate begins to slide. What is the...
Grading rubrics All the necessary equations Must write equations first, then plug values Final calculation 80% 20% A crate of eggs is located in the middle of the flatbed of a pickup truck as the truck negotiates an unbanked circular road. The radius of a circle is R = 36 m. If the coefficient of static friction between crate and truck is us = 0.4, draw free-body diagram and calculate the maximum moving speed of truck without the crate sliding.
need help with number 40 wuatis thco eent OI Knetic inctOn 38. (I) Suppose you are standing on a train accelerating at 0.20 g. What minimum coefficient of static friction must exist between your feet and the floor if you are not to slide? 39. (II) The c0efficient of static friction between hard rubber and normal street pavement is about 0.90. On how steep a hill (maximum angle) can you leave a car parked? 40. (II) A flatbed truck is...
A crate is loaded onto a flatbed truck and the coefficient of static friction between the crate and the truck is 0.45. If the driver of the truck is moving at 25 m/s and sees the car ahead of him stop, 147 m away, will the crate slide relative to the truck as the driver avoids a collision? If so, what is the acceleration (magnitude and direction) of the crate?
A crate of oranges weighing 193 N rests on a flatbed truck 2.00 m from the back of the truck. The coefficients of friction between the crate and the bed are μs = 0.900 and μk = 0.200. The truck drives on a straight, level highway at a constant 8.00 m/s. A.)What is the force of friction acting on the crate? B. )If the truck speeds up with an acceleration of 2.70 m/s2, what is the force of the friction...
Hector drives a pickup truck horizontally at 15.0 m/s. He is transporting a crate of delicate lead crystal. If the coefficient of static friction between the crate and the truck bed is 0.300, what is the minimum stopping distance for the truck so the crate will not slide? a. 51.0 m b. 38.3 m c. 44.4 m d. 28.7 m
Cumulative Problem 1 A 150.0-kg crate rests in the bed of a truck that slows from 56.0 km/h to a stop in 12.0 s. The coefficient of static friction between the crate and the truck bed is 0.675. 1) Will the crate slide during the braking period? The crate will slide during the braking period. The crate will not slide during the braking period. Submit 2) What is the minimum stopping time for the truck in order to prevent the...