Frank is pushing a block of 50 kg over the floor, with a force of 600 N downward and forward, making a 20° angle with the horizontal. The coefficient of sliding friction between the block and the floor is 0.39.
Find the acceleration of the block and the friction force acting on the block.
Let's try to sketch what is happening.
We need to draw a horizontal surface, a block on that surface, and indicate that the block is pushed by a downward-forward force that makes 20° with the horizontal. We also need to indicate that there is friction between the block and the floor:
Next, we look at our sketch, and try to list all the forces that act on the block.
There are 4 forces acting:
Knowing the forces, let's draw a free-body diagram for the block:
Let's take a step back and examine what we know, and what the need to find:
We know the mass of the block (50 kg), the push (600 N at 20° downward-forward), and the coefficient of friction (0.39).
We need to find the acceleration of the block and the force of friction.
Since we know the mass of the block, if we first find the resultant force, we can then apply Newton's 2nd Law to get the block's acceleration.
Let's start by drawing the coordinate axes on our free-body diagram, and determine the x and y components of all the forces acting on the block.
We can now find the x and y components of the resultant force:
For Eq. (1) we need to find the friction force Ff.
By definition, the direction of the sliding friction force will be opposite to the motion of the block, and its magnitude will be equal to the product between the coefficient of sliding friction and the normal force:
We know μ, but we don't know N.
Where did we see N before?
We saw N in Eq. (2):
Since the block is moving along the floor, Ry must be zero (otherwise there would be a vertical acceleration):
The only unknown in this equation is N, so we can solve it:
Having found N, we can calculate Ff:
Now that we know Ff, we can find Rx from Eq. (1):
Rx is positive (since the block accelerates in the positive x direction), and we have shown that Ry is zero. This means that R will have direction in the positive x axis, and its magnitude will be equal to Rx:
Now that we finally found the resultant force, let's apply Newton's 2nd Law to find the acceleration of the block:
|a =||F cos 20° − Ff|
|a =||(600 N) (cos 20°) − 270 N|
|a =||294 N|
And with that we have found everything we needed:
Robert is pushing a block of 61 kg over a horizontal surface. He pushes downward and forward, making an angle of 30° with the horizontal. Assuming that the push has magnitude 490 N, and that the coefficient of sliding friction between the block and the surface is 0.44, find the acceleration of the block, and the force of friction acting on the block.