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- I don't understand why Fs <= Fk. What does the F in (1) and (2) stand for? If it's force applied, wouldn't F = Fs + Fk for box 2, and (F - Fs - Fk) = Fs for box 1? My thinking is that the Force applied on box 2 minus the force of kinetic friction and the force of static friction applied by Box 1 on 2 would equal to the net force. Then the net force of box 2 would cause the same netforce on box 1, if box 1 isn't moving, then it's Fs should be equal to that net force applied on box 1. Therefore Force applied wouldn't be equal to Static Friction Force, and shouldn't it be bigger than the Kinetic Friction Force since it's moving in the direction of Force applied?Consider two peoplepushing a toboggan with four children on it up a snowcovered slope. Construct a problem in which you calculatethe acceleration of the toboggan and its load. Include a freebody diagram of the appropriate system of interest as thebasis for your analysis. Show vector forces and theircomponents and explain the choice of coordinates. Amongthe things to be considered are the forces exerted by thosepushing, the angle of the slope, and the masses of thetoboggan and children.What is the y-component of a 95.3 N force that is exerted at 57.1degree to the horizontal.
- In most of the problems you have dealt with so far, weight is equal in magnitude to that of the normal force. Is it possible for the normal force not to be equal to weight? Explain your answer. Yes, it is possible. Oftentimes, normal force counteracts the weight, thus making them equal. This happens when the object is situated on a horizontal surface and the force is exerted on the direction of the gravitational field. Otherwise, as in the case of an elevator accelerating downwards, N and W will not equalize. Yes, it is possible. Oftentimes, normal force counteracts the weight, thus making them equal. This happens when the object is situated on a vertical surface and the force is exerted on the direction of the gravitational field. Otherwise, as in the case of an elevator accelerating downwards, N and W will not equalize. No, it is not possible. Oftentimes, normal force counteracts the weight, thus making them equal. This happens when the object is situated on a horizontal…** I am confused on this being below the horizontal. Is that making Fy a negative? If so, would that then affect the calculation for normal force? A box (M = 4 kg) is resting on a frictionless horizontal surface. Then there is a 25 Newton force acting onit at an angle 35 degrees below the horizontal.1. Find the horizontal component of the force, Fx - I have Fcostheta so 25cos35 = 20.479 N2. Find the vertical component of the force, Fy - I don't know if this should be 25sin35 or 25sin(-35) 3. Find the normal force acting on the box - Fn = mg - Fy (but I do not understand if this is going to be a positive or negative Fy, obviously negative would make this mg + Fy)4. Find the acceleration of the box5. Find the distance it will travel in 3 seconds.An object of mass m has these three forces acting on it (there is no normal force, "no surface"). F1 = 1 N, F2 = 10 N, and F3 = 4 N. When answering the questions below, assume the x-direction is to the right, and the y-direction is straight upwards. What is the magnitude of the net force, in newtons? What is the angle θ, in degrees, of the net force, measured from the +x-axis? Enter an angle between -180° and 180°. What is the magnitude, |a| of the acceleration, in meters per square second, if the block has a mass of 8.9 kg?
- There are two forces on the 1.19 kg box in the overhead view of the figure but only one is shown. For F1 = 15.7 N, a = 14.7 m/s2, and θ = 34.1°, find the second force (a) in unit-vector notation and as (b) a magnitude and (c) a direction. (State the direction as a negative angle measured from the +x direction.)One paint bucket of known mass mA = 10.0kg is hanging by a massless cord from another paint bucket of known mass mB = 15.0kg, which is hanging by a massless cord. If the two buckets are pulled upward with a known acceleration of a = 2.0m/s2 by the upper cord, calculate the tension in cord 1 and the tension in cord 2 by (a) selecting appropriate common equations and making appropriate substitutions for the symbolic answer. (b) show the numeric substitution of given quantities for the final result. (c) Draw a free body diagram for each bucket.A contestant in a winter sporting event pushes an mm kg block of ice across a frozen lake by applying a force FF at an angle θθ below the horizontal as shown. Assume that the coefficient of static friction for ice on ice is 0.0300, and the coefficient of kinetic friction for the same is 0.0100. Let to the right be the positive x direction and up be the positive y direction for your equations. Obtain a numeric value, in newtons, for the magnitude of the maximum applied force, F, consistent with static friction when the force makes an angle 32° below the horizontal and the mass of the block is 63 kg. Obtain a numeric value for the acceleration, a, in meters per squared seconds, when the mass of the block is 63 kg and the angle of the rope is 32° below the horizontal.
- A 15-kg block is pushed across a rough horizontal surface by a force that is angled 30.0◦ below the horizontal. The magnitude of the force is 100 N and the acceleration of the block as it is pushed is 4 m/s2. What is the magnitude of the contact force (the vector addition of normal force and friction) exerted on the block by the surface?Consider the baby being weighed in the following figure. (a) What is the mass of the infant and basket if a scale reading of 55 N is observed? (b) What is tension T1in the cord attaching the baby to the scale? (c) What is tension T2, In the cord attaching the scale to the ceiling, If the scale has a mass of 0.500 kg? (d) Sketch the situation, indicating the system of interest used to solve each part. The masses of the cords are negligible.Drawing Free-Body Diagrams In completing the solution for a problem involving forces, what do we do after constructing the tree-body diagram? That is, what do we apply?
