A box of books weighing 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exerted downward at an angle of 35.2 degrees below the horizontal. Fink Mu,k between the box and the floor.
There have been several of this type of question posted here today. Take a look at how they were done.
Each can be solved by writing down Newton's second law. The force "F" in the law F= m a is the net force. In this case, the acceleration a is zero. Therefore the applied force component along the direction of motion equals the friction force. Write that statement in algebraic form and solve for Mu,k.
4 answers
427
1. The 425 N force needs to be broken into component vectors. 425*sin35.2 will give you the vertical component, 425*cos35.2 will give you the horizontal component.
2. The vertical component gets added to the box's 325 N weight because it is being pushed down on.
3. The horizontal component is equaled by the force of friction backward. Divide this force by the force from step 2 to get the coefficient of friction, which is about 0.61.
2. The vertical component gets added to the box's 325 N weight because it is being pushed down on.
3. The horizontal component is equaled by the force of friction backward. Divide this force by the force from step 2 to get the coefficient of friction, which is about 0.61.
20
But how did you choose the angle?
2 answers