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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
Answer:
Fn: magnitude of the net force.
Fn=30.11N , oriented 75.3 ° clockwise from the -x axis
Explanation:
Components on the x-y axes of the 17 N force(F₁)
F₁x=17*cos48°= 11.38N
F₁y=17*sin48° = 12.63 N
Components on the x-y axes of the the second force(F₂)
F₂x= −19.0 N
F₂y= 16.5 N
Components on the x-y axes of the net force (Fn)
Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N
Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N
Magnitude of the net force.
Direction of the net force (β)
β=75.3°
Magnitude and direction of the net force
Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis
In the attached graph we can observe the magnitude and direction of the net force
