#$&* course Phy 121 12/15 2 pm 026. `query 26
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Given Solution: `aIf the crate is moving then the force exerted by friction is .30 times the normal force between it and the floor. If the push is horizontal, then the only horizontal forces acting on the crate are the downward force of gravity and the upward force exerted by the floor. Since the crate is not accelerating in the vertical direction, these forces are equal and opposite so the normal force is equal to the weight of the crate. The weight of the crate is 35 kg * 9.8 m/s^2 = 340 N, approx. The frictional force is therefore f = .30 * 340 N = 100 N, approx.. If the crate moves at constant speed, then its acceleration is zero, so the net force acting on it is zero. The floor exerts its normal force upward, which counters the gravitational force (i.e., the weight). The frictional force acts in the direction opposite motion; if net force is zero an equal and opposite force is required, so you must push the box with a force of 100 N in the direction of motion. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: Suppose you have a 120-kg wooden crate resting on a wood floor. (a) What maximum force can you exert horizontally on the crate without moving it? (b) If you continue to exert this force once the crate starts to slip, what will its acceleration then be? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Well the weight of this crate would be 120 kg * 9.8 m/s^2 = 1,176 Newtons Now I am not sure what to do from here. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The coefficient of static friction for wood on wood is given in your text as 0.5, the coefficient of kinetic friction being 0.3. The weight of the crate is 120 kg * 9.8 m/s^2 = 1200 N, approx.. So you could exert any force less than or equal to 0.5 * 1200 N = 600 N without causing the crate to move. Once the crate starts to slip the coefficient of friction becomes 0.3, the coefficient of kinetic friction, and the frictional force drops to 0.3 * 1200 N = 360 N. If you continue the 600 N force, the net force will be F_net = 600 N - 360 N = 240 N. This will result in an acceleration of a = F_net / m = 240 N / 120 kg = 2 kg m/s^2 / kg = 2 m/s^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Oh I see now, I did not think to look in the text for the coefficient of the friction for wood. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: Calculate the deceleration of a snow boarder going up a 5.0º , slope assuming the coefficient of friction for waxed wood on wet snow. The result of Exercise 5.1 may be useful, but be careful to consider the fact that the snow boarder is going uphill. Explicitly show how you follow the steps in Problem-Solving Strategies. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The coefficient is 0.1. We tilt the x and y plane to where the x axis lines up with the 5 degree slope. The weight vector would therefore be going straight down. The weight vector is at 265 degrees with the positive x axis, making the x component = L * cos(265 degrees) = -0.087 * L and the y component = L * sin(265 degrees) = 0.996 * L Now I am not sure of what to do. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Assume a slope of 5 degrees, sloping up and to the right, with an x-y coordinate system rotated so that the positive x axis points up the incline. The weight vector (m g ) of the snowboarder makes an angle of 265 deg with the positive x axis, so the x and y components of the weight are (m g )_x = ( m g ) cos(265 deg) = -.1 (m g ), approx. and (m g )_y = (m g ) sin(265 deg) = .996 (m g ), approx.. (m g )_y is so close to (m g ) that we will just assume that (m g )_y = (m g ). The normal force exerted between the snowboarder and the incline is therefore (m g ), and the frictional force opposing motion is the product of the normal force (m g )_y and the coefficient of kinetic friction, which is 0.1. The frictional force acts in the direction opposite motion, which is the negative x direction, and is therefore f_frict = -0.1 (m g ) in the x direction. The x component -.1 (m g ) of the weight also opposes motion up the incline. The forces in the y direction are the normal force and the y component of the weight, which are equal and opposite, so the net force is just the total force in the x direction. We therefore have F_net = -.1 (m g ) + (-.1 (m g ) ) = -.2 (m g ). The acceleration of the snowboarder is thus a = F_net / m = (-.2 m g) / m = -.2 g, or -.2 * 9.8 m/s^2 = -2 m/s^2, approx.. This acceleration, being in the direction opposite motion, constitutes a deceleration of 2 m/s^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Now I see what I did wrong, but I am still quite confused about this. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: (a) Calculate the acceleration of a skier heading down a 10.0º slope, assuming the coefficient of friction for waxed wood on wet snow. (b) Find the angle of the slope down which this skier could coast at a constant velocity. You can neglect air resistance in both parts, and you will find the result of Exercise 5.1 to be useful. Explicitly show how you follow the steps in the Problem-Solving Strategies. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We tilt the x and y plane downward to where the x axis lines up with the -10 degree slope, which would make the weight vector to be at 280 degrees counterclockwise from the positive x axis. The x and y components would be x component = mg * cos(280) = 0.17 * mg and the y component = mg * sin(280) = -0.98 * mg The Ffrict = -0.1 * mg in the x direction Fnet = 0.17 * mg + 0.17 * mg = 0.34 * mg Therefore the a = Fnet / m = 0.34 * g or 0.34 * 9.8 m/s^2 = 3.33 m/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: A contestant in a winter sporting event pushes a 45.0-kg block of ice across a frozen lake as shown in Figure 5.23(a). (a) Calculate the minimum force F he must exert to get the block moving. (b) What is its acceleration once it starts to move, if that force is maintained? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can rotate the x and y plane down 25 degrees to make the weight component be at 295 degrees counterclockwise from the positive x axis. The weight would be 45 kg * 9.8 m/s^2 = 441 Newtons The x component would be 441 N * cos(295 degrees) = 185.22 and the y component = 441 N * sin(295 degrees) = -399.68 The coefficient is 0.3. Therefore we take the 441 Newtons * 0.3 = 132.3 Newtons Fnet = 132.3 + 441 = 573.3 Newtons a = Fnet / m = 573.3 N / 45 kg = 12.74 m/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: A contestant in a winter sporting event pushes a 45.0-kg block of ice across a frozen lake as shown in Figure 5.23(a). (a) Calculate the minimum force F he must exert to get the block moving. (b) What is its acceleration once it starts to move, if that force is maintained? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can rotate the x and y plane down 25 degrees to make the weight component be at 295 degrees counterclockwise from the positive x axis. The weight would be 45 kg * 9.8 m/s^2 = 441 Newtons The x component would be 441 N * cos(295 degrees) = 185.22 and the y component = 441 N * sin(295 degrees) = -399.68 The coefficient is 0.3. Therefore we take the 441 Newtons * 0.3 = 132.3 Newtons Fnet = 132.3 + 441 = 573.3 Newtons a = Fnet / m = 573.3 N / 45 kg = 12.74 m/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!