when finding the friction and normal force of this..
88 N is pushed on a lawn mower of 40 kg.
why do you do 88*sin45, and why isn't the normal just 140 kg? i have an example of something moving against the ground in my notes and it's simply mass times gravity.
88 N is pushed on a lawn mower of 40 kg.
why do you do 88*sin45, and why isn't the normal just 140 kg? i have an example of something moving against the ground in my notes and it's simply mass times gravity.
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It appreared from your duestion that a pull of 88 Newton is applied at 45* from the Horizontal on a 40 kg object.
This pull can be devided in to two components i.e. the horizontal component Fx = Fcos45* and the vertical component i.e Fy = Fsin45*
The Fx will try to move the object in its direction, whereas the vertical component is acting downward and thus added with the weight of the object. This result in an increased Normal reaction force
=>N = Fsin45* + mg
=>N = 88 x sin45* + 40 x 9.8
=>N = 454.23 Newton
This pull can be devided in to two components i.e. the horizontal component Fx = Fcos45* and the vertical component i.e Fy = Fsin45*
The Fx will try to move the object in its direction, whereas the vertical component is acting downward and thus added with the weight of the object. This result in an increased Normal reaction force
=>N = Fsin45* + mg
=>N = 88 x sin45* + 40 x 9.8
=>N = 454.23 Newton
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The normal force on the lawnmower if it's just sitting there with no other force applied is
N = mg = (40kg)(9.81m/s^2) = 392N
The applied force on a lawnmower is at an angle (x) from the handle. The downward component of that force is (force)*sinx. In this scenario they are giving you x=45 degrees, so the downward applied force is
F(y) = Fsinx = (88N)(sin(45)) = 62N
Total downward force on the pushed lawnmower is F(y) + N = 392N + 62N = 454N
Normal force = 454N
N = mg = (40kg)(9.81m/s^2) = 392N
The applied force on a lawnmower is at an angle (x) from the handle. The downward component of that force is (force)*sinx. In this scenario they are giving you x=45 degrees, so the downward applied force is
F(y) = Fsinx = (88N)(sin(45)) = 62N
Total downward force on the pushed lawnmower is F(y) + N = 392N + 62N = 454N
Normal force = 454N
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If I understand the question right, the 88 N is being applied 45 degrees towards the ground. That means the ground has to "push up" the weight of the mower as well as part of the force applied because the additional force has a x and y direction. The force in the y direction has to be considered as well for this. So the normal force is going to be m*g of the mower + sin45*88 N. The answer will be in Newtons by the way, not kilograms. Your other example probably doesn't have any other force pushing down on it. For the friction force, use the normal force from the above equation.