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courtrigrad
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A 30.0 kg packing case is initially at rest on the floor of a truck. The coefficient of static friction between the case and the floor is 0.30, and the coefficient of kinetic friction is 0.20. The truck is traveling due east at a constant speed. Find the magnitude and direction of the friction force acting on the case (a) when the truck accelerates at 2.20 m/s^2 eastward; (b) when it accelerates at 3.50 m/s^2 westward.
Ok so initially, we have the force of gravity [tex] w = (30 kg)(9.8 \frac{m}{s^{2}}) [/tex], the static friction force, and the normal force (equal to the weight but opposite in direction). So for parts (a) and (b) since everything is moving, we have to find the magnitude and direction of the kinetic frictional force. So [tex] f_{k} = \mu_{k}\times N [/tex]. This is where I become stuck. How do you relate the acceleration to finding the frictional force?
Any help is appreciated.
Thanks
Ok so initially, we have the force of gravity [tex] w = (30 kg)(9.8 \frac{m}{s^{2}}) [/tex], the static friction force, and the normal force (equal to the weight but opposite in direction). So for parts (a) and (b) since everything is moving, we have to find the magnitude and direction of the kinetic frictional force. So [tex] f_{k} = \mu_{k}\times N [/tex]. This is where I become stuck. How do you relate the acceleration to finding the frictional force?
Any help is appreciated.
Thanks