A force can give a certain mass a certain acceleration. if i double the force and double the mass, the acceleration of this obje
1 answer:
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Answer:
The magnitud of the force is 124.8N.
Explanation:
First we have to find the value of the static friction coefficient, when the external force F is applied to upper block (i will call it A Block) we have a free body diagram as the one shown in the figure i attached, so since this block has no aceleration in any direction the force F should be equal to the friction force between A and B block, one we noticed this we can use the equation for the Friction force to find the coefficient:
μs
and again, since the block has no acceleration the normal between A and B block should be equal to the weigth of the first block, so we have:
replacing this we have:
μs
μs*
and μs
now it's time to see the free body diagram for the b block, if we now apply the F force to the B block the diagram should look like in the figure.
the color of the arrow gives you an idea of where the force comes from, the blue ones comes from the B block, the red ones from the A block and the brown ones from the ground.
now for the B block you can see two friction forces, one for the ground and one for the A block, both of these directed bacwards, and two normal forces, again one for the ground and one for the A block but the normal force for the A block is aiming downwards.
again we use the fact that the block is not accelerating in any direction so the sum of the forces in x and y direction have to be 0, so:
This is the new external F force that we are looking for:
we know Friction2(AB) because we found that in the previous block so:
*μs
for the other friction we have to use the equation:
*μs
from y axis we have:
we found the value of Normal(AB) with the previous block so:
and:
*μs
*μs
*μs+μs*
*μs
and since: μs*
the new F force would be:
*μs
Answer:
Explanation:
Check attachment
Answer:
<h2>154.73N</h2>Explanation:
The question is incomplete. Here is the complete question.
Using the strap at an angle of 31° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15 kg school bag across the floor at a constant velocity. (a) If the force of tension in the strap is 51 N, what is the normal force.
Check the diagram related to the question in the attachment below for better understanding.
The normal force is the reaction acting perpendicular to the force of tension in the strap and opposite the weight of the bag. They are the forces acting along the vertical.
The normal force N will be the sum of the force of tension acting along the vertical (Ty) and the weight of the bag (W).
Ty = 15sin31°
Ty = 7.73N
W = mass * acceleration due to gravity
W = 15.0*9.8
W = 147N
The normal force is therefore expressed as;
N = Ty + W
N = 7.73 + 147
N = 154.73N
