A Physics / Dynamics question regarding normal force.?
A 75-kg man pushes a 35.8-kg mower starting from rest across a horizontal yard by applying a force of 75.3 N straight along the handle. The handle is inclined at an angle of 35 degrees below the horizontal. The magnitude of the mower's acceleration is 1.48 m/s^2, which lasts for 1.488 s, after which the mower moves at a constant velocity. Determine the magnitude of (a) the normal force on the mower, (b) the frictional force on the mower, (c) the maximum velocity of the mower, (d) the force needed to maintain the constant velocity.
electron12014-08-31T19:44:29Z
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Let’s use the following equations to determine the vertical and horizontal components of the force.
Vertical = F * sin θ = 75.3 * sin 35 This is the component of the man’s force that is pushing the mower downward. The normal force is equal to sum of the mower’s weight and this force. Weight = 35.8 * 9.8 = 350.84 N Normal force = 75.3 * sin 35 + 350.84 This is approximately 394 N.
Horizontal = F * cos θ = 75.3 * cos 35 Let’s use the horizontal component in the following equation to determine the acceleration without friction.
75.3 * cos 35 = 35.8 * a a = 75.3 * cos 35 ÷ 35.8
This is approximately 1.73 m/s^2. Since this is greater than the actual acceleration, let’s use the following equation to determine the friction force.
Ff = mass * (a1 – a2) Ff = 35.8 * (75.3 * cos 35 ÷ 35.8 – 1.48) = 8.698148935 N
To determine the maximum velocity, use the following equation.
vf = vi + a * t, vi = 0, a = 1.48, t = 1.488 vf = 1.48 * 1.488 = 2.20224 m/s
For the mower to move at constant velocity, the net horizontal force must be 0 N. For this to happen, the horizontal component of the force that man exerts on the handle must be equal to the friction force. As we change the force that the man exerts on the handle, the vertical and horizontal components will change. To determine the new friction force, we need to determine the coefficient of friction. Let’s use the following equation to determine the coefficient of friction.
Ff = μ * Total normal force Total normal force = 75.3 * sin 35 + 350.84
8.698148935 = μ * (75.3 * sin 35 + 350.84) μ = 8.698148935 ÷ (75.3 * sin 35 + 350.84) = 0.22074822 Let’s use the following equation to determine the magnitude of the force that the man exerts on handle.
F * cos 35 = μ * (weight + F * sin 35)
F * cos 35 = 0.22074822 * (350.84 + F * sin 35) F * cos 35 = 7.74473055 + 0.22074822 * F * sin 35 F * cos35 – 0.22074822 * F * sin 35 = 7.74473055 F = 7.74473055 ÷ (cos 35 – 0.22074822 * sin 35) = 11.18314398 N You can round as your instructor desires.
First lets resolve the applied Force into is x and y components Fx = 75.3cos35 = 61.68 N Fy = -75.3sin35 = -43.2 N ; negative because its applied downward
a) normal force on mower + weight of mower + Fy = 0 Fn = 43.2 + 35.8(9.8) = 394 N b) The acceleration of the mower = 1.48m/s^2 net force in the x direction = 35.8(1.48) = Fx - Friction 53 = 61.68 - Friction Friction = 8.68 = u(394) u = 0.022 c)Max velocity = acceleration x time = 1.48(1.488) = 2.2 m/s d)net force = Force needed in x direction - friction 0 = Fcos35 - (0.022)[(35.8)(9.8) + Fsin35] 0.82F = 7.72 + 0.0126F 0.8074F = 7.72 F = 9.56 N