Anonymous
Anonymous asked in Science & MathematicsPhysics · 3 weeks ago

# Pleas help with question below?

Bob is transporting a large stone block in the back of his pickup truck to deliver to a customer. The stone block is simply laying in the bed of the truck without any straps or rigging, and since it is too long to fit, the tailgate is left open. The customer’s driveway is a flat slope at a 15.0° upwards incline relative to horizonal. The stone block has a mass of 850.0 kg and the coefficient of static friction between the block and the bed of the pickup truck is µs=0.42 .

a)[5 pts] Draw a free body diagram for the stone block, and identify all the forces that are acting on it. Treat the bed of the pickup as a flat surface which is parallel to the driveway. Be sure to define your coordinate system (indicate the x and y directions).

b)[5 pts] Is there a maximum speed that the pickup can travel up the driveway without the stone block sliding off the bed and falling? If no, explain (and type your explanation here). If yes, calculate the maximum speed.

c)[5 pts] Is there a maximum acceleration that the pickup can have up the driveway without the stone block sliding off the bed and falling? If no, explain (and type your explanation here). If yes, calculate the maximum acceleration.

Relevance
• oubaas
Lv 7
3 weeks ago

with reference to the sketch shown below :

F// = Fe

minimum μ required for preventing block from slide :

m*g*sin A' angle = m*g*cos A' angle*μ

mass m and gravity g cancel

μ = tan A' angle

if A' angle = 15°, then tan 15° = 0.268 < 0.42 ...the block won't slide

...furthermore :

weight Fp = m*g = 850*9.806 = 8340 N

normal force FL = Fp*cos 15° = 8340*0.966 = 8050 N

reaction force R = -FL

gravit. force F// = Fp*sin 15° = 8340*0.259 = 2160 N

friction force Fe = -F//

Speed won't affect sliding , just the acceleration

Max tolerable friction force Fem = FL*μ = 8050*0.42 = 3380 N

m*a = Fem-F// = 3380-2160 = 1220 N

acceleration a = Fem/m = 1220/850 = 1.44 m/sec^2

• NCS
Lv 7
3 weeks ago

a) up the slope: a friction force equal to the downslope component of the weight:

f = m*g*sinΘ =850.0kg * 9.81m/s² * sin15.0º = 2158 N

straight down: weight, = m*g = 8339 N

perpendicular to and out of the truck bed: a normal force

Fn = m*g*cosΘ = 8054 N

b) There is no maximum speed. Any constant speed constitutes equilibrium. It's the acceleration that matters.

c) Parallel to the incline,

net force = maximum friction force - weight component

m*a = µs*m*g*cosΘ - m*g*sinΘ

mass m cancels

max a = 9.81m/s² * (0.42*cos15.0º - sin15.0º) = 1.44 m/s²